Pyth, 27 26 28 bytes

-^2le+k}#"lucky".:Q)yl@"omen

1 byte saved thanks to OP :-)

Explanation:

                                 Implicit Q as input
                .:Q              Find all substrings of input
     +k}#"lucky"                 Filter for substring of "lucky", prepend "" in case of []
    e                            Take last element, which is longest
   l                             Get its length
 ^2                              Raise two to that
                      @"omen"Q   Filter Q for characters in "omen"
                     l           Get length; counts how many characters in "omen" there are
                    y            Double that
-                                Find the difference

Test it here.

Ruby, 91 87 bytes

String#count's finnicky usage strikes again! (When passed a String, it counts all occurrences of each letter in the function argument instead of all occurrences of the entire string.)

Try it online

->s{2**(z=0..5).max_by{|j|z.map{|i|s[b="lucky"[i,j]]?b.size: 0}.max}-2*s.count("omen")}

A version that takes in lines from STDIN and prints them: 89 bytes (86 +3 from the -n flag)

p 2**(z=0..5).max_by{|j|z.map{|i|$_[b="lucky"[i,j]]?b.size: 0}.max}-2*$_.count("omen")

Ruby: 100 bytes

->s{2**(m=0;4.times{|j|1.upto(5){|i|m=[i,m].max if s.match"lucky"[j,i]}};m)-s.scan(/[omen]/).size*2}

JavaScript (ES7), 123 112 107 bytes

s=>2**[5,4,3,2,1,0].find((i,_,a)=>a.some(j=>s.includes("luckyL".substr(j,i))))-2*~-s.split(/[omen]/).length

Edit: Saved 11 bytes thanks to @Titus by assuming that the letter L does not appear in the input. Saved 5 bytes thanks to @Oriol. ES6 version for 125 114 109 bytes:

f=
s=>(1<<[5,4,3,2,1,0].find((i,_,a)=>a.some(j=>s.includes("luckyL".substr(j,i)))))-2*~-s.split(/[omen]/).length
;

<input oninput=o.textContent=f(this.value)><pre id=o></pre>

Javascript - 206 Bytes

r=>{var a="lucky";r:for(var e=5;e>0;e--)for(var n=0;6>n+e;n++){var o=a.substring(n,e+n);if(r.includes(o))break r}for(var t=0,e=0;e<r.length;e++)('omen'.indexOf(r[e])+1)&&t++;return Math.pow(2,o.length)-2*t}

hashmap, 38 bytes (non-competing)

Non-competing, however, because hashmap can't do consecutive orders yet.

i&sl"[^lucky]"-L2#^:n"[^omen]"-L2*;n#-

Explanation:

i&sl"[^lucky]"-L2#^:n"[^omen]"-L2*;n#-
i&sl                                   Get input, duplicate, lowercase
    "[^lucky]"-                        Remove all characters that are not in "lucky"
               L2#^                    Use the POW! command
                   :n                  Add to variable n
                     "[^omen]"-        Remove all characters that are not in "omen"
                               L2*     Multiply the length by 2
                                  ;n#- Get variable n and subtract

Python (139 Bytes)

import itertools as t
s=input()
print 2**max([j-i for i,j in t.combinations(range(6),2)if'lucky'[i:j]in s]+[0])-2*sum(_ in'omen'for _ in s)

Ruby, 57 bytes

b=gets.count'omen'
$.+=1while/[lucky]{#$.}/
p 2**$./2-2*b

gets sets $. to 1 as a side effect, then we increment it until the regular expression matching $. consecutive lucky characters no longer matches.

TSQL, 233 bytes

Golfed:

DECLARE @t varchar(99)='oluck'

,@z INT=0,@a INT=0,@ INT=1,@c INT=0WHILE @a<LEN(@t)SELECT
@a+=IIF(@=1,1,0),@z=IIF('LUCKY'LIKE'%'+x+'%'and @>@z,@,@z),@c+=IIF(x
IN('O','M','E','N'),2,0),@=IIF(@+@a-1=LEN(@t),1,@+1)FROM(SELECT
SUBSTRING(@t,@a,@)x)x PRINT POWER(2,@z)-@c

Ungolfed:

DECLARE @t varchar(99)='oluck'

,@z INT=0
,@a INT=0
,@  INT=1
,@c INT=0
WHILE @a<LEN(@t)
  SELECT
    @a+=IIF(@=1,1,0),
    @z=IIF('LUCKY'LIKE'%'+x+'%'and @>@z,@,@z),
    @c+=IIF(x IN('O','M','E','N'),2,0),
    @=IIF(@+@a-1=LEN(@t),1,@+1)
    FROM(SELECT SUBSTRING(@t,@a,@)x)x
PRINT POWER(2,@z)-@c

Try it online

Mathematica, 86 bytes

Code:

2^StringLength@LongestCommonSubsequence[#,"lucky"]-2StringCount[#,{"o","m","e","n"}]&

Explanation:

LongestCommonSubsequence returns the longest contiguous substring common to the input and "lucky". StringLength gives its length. StringCount counts the number of occurrences of the characters of "omen" in the input.

Haskell (134 132 Bytes)

import Data.List
c[]=[]
c s@(_:x)=inits s++c x
l=length
g q=2^(maximum$map(\z->l q-l(q\\z))$c"lucky")-2*(l$intersect q"omen")

Not a code golfer nor a Haskell programmer, so would love some tips on this one.

(Example: g "firetruck")

Haskell, 99

Another approach... I just learned about function aliasing

import Data.List
s=subsequences
i=intersect
l=length
f n=2^(l$last$i(s"lucky")$s n)-2*l(i n$"omen")

Haskell is my new love :D

Usage

f"lucky"
32

f"firetruck"
6

f"memes"
-7

Python 3, 168 157 152 139 144 136 bytes

EDIT: Really obvious things I should have seen easier have been changed, and some slightly less obvious.

Edit 2: stoopid (˚n˚). The program threw errors. I fixed it up. not actually 153 :(

Thanks to Leaky Nun for saving 5 bytes, and to jmilloy for saving 13 8 bytes.

s=input()
p=q=k=len(s)
m=0
while-~p:
 while-~q:m=(m,q-p)[(s[p:q]in"lucky")*q-p>m];q-=1
 p-=1;q=k
print(2**m-2*sum(i in"omen"for i in s))

The program runs through all possibly possible substrings in input (possibly possible, because it computes impossible substrings as well, 8 to 7, for example), checks if the substring is in "lucky", then sets the exponent of 2 to the length of the substring should it be greater than the current value. Possibly could be improved by using only one while loop. Could possibly use some improvement; I'm still getting the hang of this.

PHP program, 139 135 108 bytes

quantum leap fails for multiple substrings where the first occurence is shorter. :(

^{actually I could save another 7 bytes in PHP<5.4 with register_globals on}

<?for($s=$argv[1];$i<5;$i++)for($j=6;--$j;)$r=max($r,strstr($s,substr('lucky*',$i,$j))?2**$j:1);echo$r-2*preg_match_all('/[omen]/',$s);

usage: php -d error_reporting=0 <filename> <string>

+5 for a function:

function f($s){for(;$i<5;$i++)for($j=6;--$j;)$r=max($r,strstr($s,substr('lucky*',$i,$j))?2**$j:1);return$r-2*preg_match_all('/[omen]/',$s);}

tests (on the function)

echo '<table border=1><tr><th>input</th><th>output</th><th>expected</th><th>ok?</th></tr>';
foreach([
    'lumberjack'=>0,        'smack'=>2,
    'nnnnnomemenn'=>-23,    'lucky'=>32,
    'omen'=>-7,             'firetruck'=>6,
    'memes'=>-7,            'determine the “luck” of a string'=>0,
    'amazing'=>-3,          'wicked'=>2,
    'chucky'=>16,           'uckyuke'=>14,
    'ugly'=>2,              'lucy'=>8,
    'lukelucky'=>30
] as $x=>$e){
    $y=f($x);
    echo"$h<tr><td>",$x,'</td><td>',$y,'</td><td>',$e,'</td><td>',$e==$y?'Y':'N',"</td></tr>";
}echo '</table>';