Python 3, 40 bytes

lambda a:sum(map(int,map(str,a),[10]+a))

Tests are at ideone

map(str, a) creates a generator, G, that calls str on each value in a, converting to strings
map(int, G, [10]+a) creates a generator that calls int(g, v) for pairs across G and [10]+a
int(g, v) converts the string g from the integer base v (if v is in [2,36] and g is valid)
sum does what it says on the tin

Python 2, 48 bytes

lambda a:sum(int(`x`,y)for x,y in zip(a,[10]+a))

Tests are at ideone

zip(a,[10]+a) traverses pairs of the values in a, and the previous value or 10 for the first
the backticks in the int call convert x to a string, s
int(s, y) converts the string s from the integer base y (if y is in [2,36] and s is valid)
sum does what it says on the tin

PHP, 53 51 bytes

for(;$x=$argv[++$i];$b=$x)$s+=intval($x,$b);echo$s;

Iterates over the input, converting each input to string variant. Then takes the integer value using the previous number as base. For the first number, the base will not be set, PHP will then start off with 10 (inferred from the number format).

Run like this (-d added for aesthetics only):

php -d error_reporting=30709 -r 'for(;$x=$argv[++$i];$b=$x)$s+=intval($x,$b);echo$s;' -- 12 11 10 9 8 7 6 5 4 3 12 2 11 3 10 2 10;echo

Tweaks

• Actually, no need to convert to string, as CLI arguments are already string. Saved 2 bytes.

Perl, 35 34 33 bytes

Includes +2 for -ap

Run with the list of numbers on STDIN:

basemix.pl <<< "4 12 34 20 14 6 25 13 33";echo

basemix.pl:

#!/usr/bin/perl -ap
$\+=$&+"$`$& 10"*/.$/*$`for@F}{

I've been waiting for ages for a chance to use this abuse...

Explanation

The input numbers can have at most 2 digits. A number xy in base b is simply b*x+y. I'm going to use the regex /.$/ so the first digit ends up in $` and the last digit in $&, so the contribution to the sum is $&+$b*$`.

I abuse the fact that for does not properly localize the regex variables (like for example map and while do) so the results of a match in the previous loop is still available in the current loop. So if I'm careful about the order in which I do the operations the base is available as "$`$&", except for the very first loop where I need the base to be 10. So I use "$`$& 10" instead

The way the first $& works is an abuse too since it is actually changed by the /.$/ while it is already on the stack waiting to be added.

The final abuse is the }{ at the end which changes the loop implied by -p from

LINE: while (defined($_ = <ARGV>)) {
    ...code..
}
continue {
    die "-p destination: $!\n" unless print $_;
}

to

LINE: while (defined($_ = <ARGV>)) {
    ...code..
}
{
}
continue {
    die "-p destination: $!\n" unless print $_;
}

Which means $_ will be undefined in the print, but it still adds $\ in which I accumulated the sum. It is also a standard golf trick to get post-loop processing

Jelly, 7 bytes

ṙ1DḅṖḌS

Try it online!

05AB1E, 7 6 bytes

Uses CP-1252 encoding.

T¸ìüöO

Explanation

Input list is taken reversed, as allowed by challenge specification.

T¸ì     # append a 10 to the list
   üö   # reduce by conversion to base-10
     O  # sum

Try it online!

Modified testsuite

Java 7, 109 89 bytes

int c(int[]a){for(Integer i=1;i<a.length;a[0]+=i.valueOf(a[i]+"",a[++i-2]));return a[0];}

Golfed 20 bytes thanks to @cliffroot (of which 12 because of a stupid mistake I made myself).

Ungolfed & test code:

Try it here.

class M{
  static int c(int[] a){
     for(Integer i = 1; i < a.length; a[0] += i.valueOf(a[i]+"", a[++i-2]));
     return a[0];
  }

  public static void main(String[] a){
    System.out.println(c(new int[]{ 4, 12, 34, 20, 14, 6, 25, 13, 33 }));
    System.out.println(c(new int[]{ 5, 14, 2, 11, 30, 18 }));
    System.out.println(c(new int[]{ 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 12, 2, 11, 3, 10, 2, 10 }));
    System.out.println(c(new int[]{ 36, 36 }));
  }
}

Output:

235
90
98
150

Java, 86 bytes

s->{int[]b={10};return s.reduce(0,(r,n)->{r+=n.valueOf(""+n,b[0]);b[0]=n;return r;});}

Testing and ungolfed

import java.util.function.ToIntFunction;
import java.util.stream.Stream;

public class Main {

  public static void main(String[] args) {
    ToIntFunction<Stream<Integer>> f = s -> {
      int[] b = {10};                 // Base, initialized with 10
      return s.reduce(0, (r, n) -> {  // Typical use of reduction, sum starts with 0.
        r += n.valueOf("" + n, b[0]); // Add the value in the previous base.
        b[0] = n;                     // Assign the new base;
        return r;
      });
    };

    System.out.println(f.applyAsInt(Stream.of(new Integer[]{4, 12, 34, 20, 14, 6, 25, 13, 33})));
  }
}

JavaScript ES6, 45 42 41 Bytes

const g =
     a=>a.map(v=>s+=parseInt(v,p,p=v),s=p=0)|s
;

console.log(g.toString().length);                                            // 42
console.log(g([4, 12, 34, 20, 14, 6, 25, 13, 33]));                          // 235
console.log(g([5, 14, 2, 11, 30, 18]  ));                                    // 90
console.log(g([12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 12, 2, 11, 3, 10, 2, 10] )); // 98

Conventiently parseInt(x,0) === parseInt(x,10).

edit: Saved 1 byte thanks to @ETHproductions

Pure bash, 38

b=10
for i;{((t+=$b#$i,b=i));}
echo $t

The input list is given at the command-line. for i; automatically iterates over the input parameters (equivalent to for i in $@;).

Ideone.

Actually, 12 bytes

;p(dX(♂$♀¿Σ+

Try it online!

Explanation:

;p(dX(♂$♀¿Σ+
;             dupe input
 p            pop first element from list
  (dX         pop and discard last element from other copy
     (♂$      stringify all elements in first copy
        ♀¿    for each pair of elements in the two lists, interpret the first element as a base-(second element) integer
          Σ   sum
           +  add first element of original list

CJam, 15 bytes

l~{A\:A10bb}%:+

Try it online!

Explanation

l~     e# Read and evaluate input.
{      e# Map this block over the input...
  A    e#   Push A. Initially this is 10, afterwards it will be the value of the
       e#   last iteration.
  \:A  e#   Swap with current value and store that in A for the next iteration.
  10b  e#   Convert to base 10 to get its decimal digits.
  b    e#   Interpret those in the base of the previous A.
}%
:+     e# Sum all of those values.

JavaScript (ES6), 54 48 bytes

I used a recursive approach.

f=(a,b)=>a[0]?parseInt(a,b)+f(a.slice(1),a[0]):0

Saved 6 bytes, thanks to Lmis!

Haskell, 65 59 bytes

b%x|x<1=x|y<-div x 10=b*b%y+x-10*y
sum.((10:)>>=zipWith(%))

Test it on Ideone.

Matlab, 68 bytes

Not a very creative solution, but here it is:

function[s]=r(x);j=10;s=0;for(i=x)s=s+base2dec(num2str(i),j);j=i;end

Tests:

>> r([4,12,34,20,14,6,25,13,33])
ans =
   235
>> r([12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 12, 2, 11, 3, 10, 2, 10])
ans =
   98
>> r([5, 14, 2, 11, 30, 18])
ans =
   90
>> r([36,36])
ans =
   150

Python 2, 52 bytes

f=lambda x:x[1:]and int(`x.pop()`,x[-1])+f(x)or x[0]

Test it on Ideone.

Julia, 63 Bytes

f(l)=sum([parse(Int,string(l[i]),l[i-1])for i=2:length(l)])+l[]

Parses each number (except the first) taking the previous element as base and sums. Adds the first element at the end

Ruby, 52 bytes

->a{eval a.zip([10]+a).map{|e|'"%s".to_i(%s)'%e}*?+}

ungolfed

->a{
  eval(
    a.zip([10]+a).map { |e|
      '"%s".to_i(%s)' % e
    }.join("+")
  )
}

usage

f=->a{eval a.zip([10]+a).map{|e|'"%s".to_i(%s)'%e}*?+}
p f[[4, 12, 34, 20, 14, 6, 25, 13, 33]] # => 235

Scala, 67 bytes

def f(a:Int*)=a zip(10+:a)map{t=>Integer.parseInt(""+t._1,t._2)}sum

Explanation:

def f(a: Int*) =     //declare a method f with varargs of type Int as parameter
a zip (10 +: a)      //zip a with 10 prepended to a, resulting in...
                     //...Array((4,10), (12,4), (34,12), (20,34), (14,20), (6,14), (25,6), (13,25), (33,13))
map { t =>           //map each tuple t to...
  Integer.parseInt(  //...an integer by parsing...
    ""+t._1, t._2    //...a string of the first item in base-second-item.
  )
}
sum                  //and sum

Mathematica, 59 bytes

I wish Mathematica's function names were shorter. But otherwise I'm happy.

Tr[FromDigits@@@Transpose@{IntegerDigits/@{##,0},{10,##}}]&

For example,

Tr[FromDigits@@@Transpose@{IntegerDigits/@{##,0},{10,##}}]&[4,12,34,20,14,6,25,13,33]

yields 235.

{##,0} is a list of the input arguments with 0 appended (representing the numerals); {10,##} is a list of the input arguments with 10 prepended (representing the bases). That pair of lists is Transposed to associate each with numeral with its base, and FromDigits (yay!) converts each numeral-base pair to a base-10 integer, the results of which are summed by Tr.