Get count of digits of a number that divide the number
Do one thing

Let's start with your top level:

def count_div_digits(l):
    for number in l:


Firstly, l is a terrible variable name. It looks like 1. Avoid it at all costs. Secondly, this function is less useful than it could be simply because it requires a list of numbers. It'd be better if the function took a single number, then you could call it on a list separately if you want:

def count_div_digits(number):
    ...

for number in lst:
    count_div_digits(number)


Return, don't print

Right now, your count_div_digits simply prints its result. That's all well and good, but it'd be better if you just returned the number. If you want to print it, you can always just do:

print count_div_digits(number)


And this way, you can store the results somewhere else for further analysis later.

Avoid continue:

The less weird hops your loops make, the easier it is to follow along the logic. Right now you have:

if digit=='0':
    continue
else:    
    # stuff


But it'd be clearer to just do:

if digit != '0':
    # same stuff


Proposed Solution

def count_div_digits(number):
    count = 0
    for digit in str(number):
        if digit != '0' and number % int(digit) == 0:
            count += 1
    return count


Generator Expressions

You could also write the above with the help of a generator expression by factoring out a length:

def sizeof(iterable):
    return sum(1 for _ in iterable)

def count_div_digits(number):
    return sizeof(digit for digit in str(number)
                  if digit != '0' and number % int(digit) == 0)


Whether you consider that an improvement or not is a matter of opinion.