Create iterator to return elements from each iterable one by one

You can use itertools.chain.from_iterable() to flatten the sequence, and use a generator expression to filter out the None values:

from itertools import chain, zip_longest

it = (v for v in chain.from_iterable(zip_longest(l1, l2)) if v is not None)

Rather than use None as the sentinel value, you may want to use a dedicated sentinel so you can use None in the input list:

_sentinel = object()
flattened = chain.from_iterable(zip_longest(l1, l2, fillvalue=_sentinel))
it = (v for v in flattened if v is not _sentinel)

If you want to filter out falsey values, then you can also use filter(None, ...):

it = filter(None, chain.from_iterable(zip_longest(l1, l2)))

Demo:

>>> from itertools import chain, zip_longest
>>> l1 = [1, 2, 3, 4, 5, 6]
>>> l2 = ['a', 'b', 'c', 'd']
>>> it = (v for v in chain.from_iterable(zip_longest(l1, l2)) if v is not None)
>>> list(it)
[1, 'a', 2, 'b', 3, 'c', 4, 'd', 5, 6]

and with a local sentinel:

>>> l1 = [1, None, 2, None, 3, None]
>>> l2 = ['a', 'b', 'c', 'd']
>>> _sentinel = object()
>>> flattened = chain.from_iterable(zip_longest(l1, l2, fillvalue=_sentinel))
>>> it = (v for v in flattened if v is not _sentinel)
>>> list(it)
[1, 'a', None, 'b', 2, 'c', None, 'd', 3, None]

The itertools recipes section also has:

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

If you want a modified version of your code, building a generator from the start (no storing list l):

import itertools
l1=[1,2,3,4,5,6]
l2=['a','b','c','d']

def flat_zip(l1,l2):
    for x,y in itertools.zip_longest(l1,l2):
        if x:
            yield x
        if y:
            yield y
it=flat_zip(l1,l2)

Though I advise for using the builtin solutions above.