Skip multiple iterations in loop

for uses iter(song) to loop; you can do this in your own code and then advance the iterator inside the loop; calling iter() on the iterable again will only return the same iterable object so you can advance the iterable inside the loop with for following right along in the next iteration.

Advance the iterator with the next() function; it works correctly in both Python 2 and 3 without having to adjust syntax:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter)
        next(song_iter)
        next(song_iter)
        print 'a' + next(song_iter)

By moving the print sing line up we can avoid repeating ourselves too.

Using next() this way can raise a StopIteration exception, if the iterable is out of values.

You could catch that exception, but it'd be easier to give next() a second argument, a default value to ignore the exception and return the default instead:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter, None)
        next(song_iter, None)
        next(song_iter, None)
        print 'a' + next(song_iter, '')

I'd use itertools.islice() to skip 3 elements instead; saves repeated next() calls:

from itertools import islice

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        print 'a' + next(islice(song_iter, 3, 4), '')

The islice(song_iter, 3, 4) iterable will skip 3 elements, then return the 4th, then be done. Calling next() on that object thus retrieves the 4th element from song_iter().

Demo:

>>> from itertools import islice
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> song_iter = iter(song)
>>> for sing in song_iter:
...     print sing
...     if sing == 'look':
...         print 'a' + next(islice(song_iter, 3, 4), '')
... 
always
look
aside
of
life

>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> count = 0
>>> while count < (len(song)):
    if song[count] == "look" :
        print song[count]
        count += 4
        song[count] = 'a' + song[count]
        continue
    print song[count]
    count += 1

Output:

always
look
aside
of
life

I think, it's just fine to use iterators and next here:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
it = iter(song)
while True:
    word = next(it, None)
    if not word:
       break
    print word
    if word == 'look':
        for _ in range(4): # skip 3 and take 4th
            word = next(it, None)
        if word:
            print 'a' + word

or, with exception handling (which is shorter as well as more robust as @Steinar noticed):

it = iter(song)
while True:
    try:
        word = next(it)
        print word
        if word == 'look':
            for _ in range(4):
                word = next(it)
            print 'a' + word 
    except StopIteration:
        break

Skip multiple iterations in loop

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