Python 2, 41 bytes

Saved 3 bytes thanks to @xnor

a,b="ab";exec"print a;a,b=b,a+b;"*input()

Test on Ideone

Simply follows the recursive definition.

Haskell, 29 35 32 bytes

a%b=a:b%(a++b)
(`take`("a"%"b"))

Simple recursion.

For reference: the old version (an adaption of this answer), concatenated the strings in the wrong order, so I had to add a flip(...) which made it too long (35 bytes).

f="a":scanl(flip(++))"b"f
(`take`f)

05AB1E, 12 11 bytes

Thanks to Emigna for saving a byte!

'a='b¹GD=Š«

Uses the CP-1252 encoding. Try it online!

Jelly, 11 10 bytes

”a”bṄ;¥@¡f

Try it online!

How it works

”a”bṄ;¥@¡f  Main link. Argument: n

”a          Set the return value to 'a'.
  ”b    ¡   Call the link to the left n times, with left argument 'b' and right
            argument 'a'. After each call, the right argument is replaced with the
            left one, and the left argument with the return value. The final
            return value is yielded by the quicklink.
      ¥       Combine the two atoms to the left into a dyadic chain.
    Ṅ           Print the left argument of the chain, followed by a linefeed.
     ;          Concatenate the left and right argument of the chain.
       @      Call the chain with reversed argument order.
         f  Filter the result by presence in n. This yields an empty string
            and thus suppresses the implicit output.

Java 7, 69 bytes

String c(int n,String a,String b){return n--<1?"":a+"\n"+c(n,b,a+b);}

ungolfed

 class fibb {


public static void main(String[] args) {
    System.out.println( c( 7, "a" , "b" ) );

}
static String c(int n,String a,String b) {

    return n-- < 1  ? "" : a + "\n" + c(n,b,a + b);

}
}

JavaScript (ES6), 43 42 bytes

Saved a byte thanks to @Arnauld

f=(n,a="a",b="b")=>n&&f(n-!alert(a),b,a+b)

Emacs, 26, 25-ish keystrokes

Program

#n to be read as key with digit(s) n:

ARETBRETF3UPUPC-SPACEC-EM-WDOWNDOWNC-Y UPC-AC-SPACEC-EM-WDOWNC-EC-YRETF4C-#(n-2)F4

Explanation

command(s)               explanation                      buffer reads (| = cursor aka point)
-----------------------------------------------------------------------------------------------
A<RET> B<RET>            input starting points            "a\nb\n|"
<F3>                     start new macro                  "a\nb\n|"
<UP><UP>                 move point two lines up          "|a\nb\n"
C-<SPACE> C-E M-W        copy line at point               "a|\nb\n"
<DOWN><DOWN>             move point two lines down        "a\nb\n|"
C-Y                      yank (paste)                     "a\nb\na|"
<UP>                     move point one line up           "a\nb|\na"
C-A C-<SPACE> C-E M-W    copy line at point               "a\nb|\na"
<DOWN>                   move point one line down         "a\nb|\na|"
C-E C-Y <RET>            yank (paste) and add new line    "a\nb|\nab\n|"
<F4>                     stop macro recording             "a\nb|\nab\n|"
C-#(n-3) <F4>            apply macro n-3 times            "a\nb|\nab\nbab\nabbab\n|"

With n=10

a
b
ab
bab
abbab
bababbab
abbabbababbab
bababbababbabbababbab
abbabbababbabbababbababbabbababbab
bababbababbabbababbababbabbababbabbababbababbabbababbab

CJam, 19 17 bytes

'a'b{_@_n\+}ri*;;

explanation

"a": Push character literal "a" onto the stack.
"b": Push character literal "b" onto the stack.
{_@_p\+}
    {: Block begin.
    _: duplicate top element on the stack
    @: rotate top 3 elements on the stack
    _: duplicate top element on the stack
    n: print string representation
    \: swap top 2 elements on the stack
    +: add, concat
    }: Block end.
r: read token (whitespace-separated)
i: convert to integer
*: multiply, join, repeat, fold (reduce)
;: pop and discard
;: pop and discard

V, 18 bytes

ia
bkÀñyjGpgJkñdj

Try it online!

Or, the more readable version:

ia
b<esc>kÀñyjGpgJkñdj

Explanation:

ia
b<esc>          " Insert the starting text and escape back to normal mode
k               " Move up a line
 Àñ       ñ     " Arg1 times:
   yj           "   Yank the current line and the line below
     G          "   Move to the end of the buffer
      p         "   Paste what we just yanked
       gJ       "   Join these two lines
         k      "   Move up one line
           dj   " Delete the last two lines

MATL, 14 bytes

97c98ci2-:"yyh

Try it online!

97c     % Push 'a'
98c     % Push 'b'
i2-     % Input number. Subtract 2
:"      % Repeat that many times
  yy    %   Duplicate the top two elements
  h     %   Concatenate them

Python 2, 55 bytes

def f(n):m='a','b';exec'print m[-2];m+=m[-2]+m[-1],;'*n

Retina, 33 bytes

.+
$*
^11
a¶b
+`¶(.+?)1
¶$1¶$%`$1

Try it online!

Saved 10 (!) bytes thanks to @MartinEnder!

Explanation

Converts the input to unary, subtracts 2 and adds the a and the b, then recursively replaces the remaining 1s with the concatenation of the two previous strings.

Batch, 102 93 bytes

@set a=a
@set b=b
@for /l %%i in (2,1,%1)do @call:l
:l
@echo %a%
@set a=%b%&set b=%a%%b%

Fortunately variables are expanded for every line before assignments take effect, so I can set both a and b using their old values without needing a temporary. Edit: Saved 9 bytes thanks to @nephi12.

Swift 3, 76 Bytes

func f(_ n:Int,_ x:String="a",_ y:String="b"){if n>1{print(x);f(n-1,y,x+y)}}

Perl, 45 +1 = 46 bytes

+1 byte for -n flag

$a=a,$b=b;say($a),($a,$b)=($b,$a.$b)for 1..$_

Slight improvement over the existing 49-byte solution, but developed separately. The parentheses for say($a) are necessary because otherwise, it interprets $a,($a,$b)=($b,$a.$b) as the argument of say which outputs more junk than we need.

Perl, 48 bytes

47 bytes code + 1 for -n.

Simple approach. Try using an array slice originally $a[@a]="@a[-2,-1]" but that necessitates $"="" or similar :(. Save 1 byte thanks to @Dada!

@;=(a,b);$;[@;]=$;[-2].$;[-1]for 3..$_;say for@

Usage

perl -nE '@;=(a,b);$;[@;]=$;[-2].$;[-1]for 3..$_;say for@' <<< 5
a
b
ab
bab
abbab

Stack my Golf, 63 bytes

Get my language here: https://github.com/cheertarts/Stack-My-Golf.

($1=(_(a))($2=(_(b))($2-f:1-f,)?)?)f\($1=(f)($1-p(\n),:f,)?)p\p

There is probably a shorter way but this is the most obvious.

Pyth, 17 bytes

J,\a\bjP.U=+Js>2J

A program that takes input of an integer and prints the result.

Try it online!

How it works

J,\a\bjP.U=+Js>2J  Program. Input: Q
 ,\a\b             Yield the two element list ['a', 'b']
J                  Assign to J
        .U         Reduce over [0, 1, 2, 3, ..., Q] (implicit input):
          =J+        J = J +
              >2J     the last two elements of J
             s        concatenated
       P           All of that except the last element
      j            Join on newlines
                   Implicitly print

Pyth - 16 15 bytes

A"ab" m=H+
~GHH

Try it online here.

APL, 30 bytes.

⎕IO must be 1.

{⎕←⍵}¨{⍵∊⍳2:⍵⌷'ab'⋄∊∇¨⍵-⌽⍳2}¨⍳

Mathematica, 49 bytes

f@1="a";f@2="b";f@n_:=f[n-2]<>f[n-1];g=f~Array~#&

Defines a function g taking the single numerical input; returns a list of strings. Straightforward recursive implementation, using the string-joining operator <>.

Mathematica, 56 bytes

NestList[#~StringReplace~{"a"->"b","b"->"ab"}&,"a",#-1]&

Unnamed function, same input/output format as above. This solution uses an alternate way to generate the strings: each string in the list is the result of simultaneously replacing, in the previous string, all occurrences of "a" with "b" and all occurrences of "b" with "ab".

PHP, 63 bytes

Recursive version:

function f($n,$a=a,$b=b){return $n--?"$a
".f($n,$b,$a.$b):'';}

Groovy, 79 Bytes

{println("a\nb");x=['a','b'];(it-2).times{println(y=x.sum());x[0]=x[1];x[1]=y}}

PHP, 53 bytes

for($a=b,$b=a;$argv[1]--;$a=($_=$b).$b=$a)echo$b.'
';

C, 168 bytes

#include<stdio.h>
int main(){char u[999]="a",v[999]="b",*a=u,*b=a+1,*c=v,*d=c+1,*e,i;for(i=0;i<5;++i){printf("%s\n",a);for(e=c;*e;*b++=*e++);e=a;a=c;c=e;e=b;b=d;d=e;}}

Two buffers (u & v) store the last two lines. The newest line (tracked with two pointers: start=c, end=d) is appended to the oldest one (start=a, end=b). Swap (a,b) and (c,d), and loop. Pay attention to buffer size before requesting too much lines. Not so short (as expected of a low-level language), but was fun to code.

C++11, 89 98 bytes

+7 bytes for all lines, not only the last one. +2 bytes more for N being the number of lines printed, not some 0-based stuff.

#include<string>
using S=std::string;S f(int n,S a="a",S b="b"){return n-1?a+"\n"+f(n-1,b,a+b):a;}

Usage:

f(5)

PHP, 61 bytes

<?=$a=a;for($b=b;--$argv[1];$b=$c.$b){$c=$a;$a=$b;echo"
$b";}

come php 7.1 you can save 3 bytes by changing to:

<?=$a=a;for($b=b;--$argv[1];[$a,$b]=[$b,$a.$b])echo"
$b";

If a trailing new line is acceptable then:

for($b=b;$argv[1]--;$b=$c.$b){$c=$a?:a;$a=$b;echo"$c
";}

is only 56 bytes.

Ruby (1.9+) 46 bytes

a,b=?a,?b;ARGV[0].to_i.times{puts a;a,b=b,a+b}

Pyth - 14 bytes

(yeah, another pyth answer)

A<G2VQG=H+~GHH

Try it online

Explanation

A<G2              - take "ab" from the var G, dump "a" into G and "b" into H.
    VQ            - for N in range(Q)
      G           - print G
          ~GH     - like =GH but returns the previous G
         +   H    - concatenate the old G with the current H
       =H         - put that value back into H

Actually, 16 bytes

There might be a shorter way to duplicate the top two elements of the stack in Actually, but I'm not sure what that way is. Golfing suggestions welcome. Try it online!

'a'b,¬`2╟2αio`na

Ungolfing

'a'b     Push "a" then "b" to the stack.
,        Explicitly take input n.
¬        Push n-2.
`...`n   Run the following function n-2 times.
  2╟2α     Push a list that repeats the top two elements of the stack twice.
  i        Flatten that list to the stack. Stack: b, a, b, a, prev_iterations
  o        Append b to the end of a. Stack: ab, b, a, prev_iterations
a        Reverse the stack so that the strings will print in the correct order.
         Implicit print all of the elements in the stack.

A different approach

This is also 16 bytes. Try it online!

'a'b,¬`│ok╔Ri`na

Ungolfing

The main difference is the function in the middle, so I'll only ungolf that here.

`...`n   Run the following function n-2 times.
  │        Duplicate the stack.
  o        Append the last item to the end of the second-to-last item.
  k╔       Uniquify the stack by wrapping the stack in a list and then calling uniquify().
  Ri       Reverse that list before returning everything to the stack
            so that the last items are at TOS again.