Ideally you would use str.find or str.index like demented hedgehog said. But you said you can't ...

Your problem is your code searches only for the first character of your search string which(the first one) is at index 2.

You are basically saying if char[0] is in s, increment index until ch == char[0] which returned 3 when I tested it but it was still wrong. Here's a way to do it.

def find_str(s, char):
    index = 0

    if char in s:
        c = char[0]
        for ch in s:
            if ch == c:
                if s[index:index+len(char)] == char:
                    return index

            index += 1

    return -1

print(find_str("Happy birthday", "py"))
print(find_str("Happy birthday", "rth"))
print(find_str("Happy birthday", "rh"))

It produced the following output:

3
8
-1

>>> val = 's2*20'
>>> val.split('*')[1]
'20'

Depending on what you want, you may want to check for '*' inside your string, or just taking everything after the first occurrence of '*', but hard to guess without more info. This might be a plausible scenario:

>>> def rest(s):
...     return s.split('*',1) if '*' in s else s
...
>>> rest('hi')
'hi'
>>> rest('hi there * wazzup * man')
['hi there ', ' wazzup * man']

Edit: as pointed out in the comment by @Jon (no, not Skeet), using partition is better in every way.

>>> val = 's2*20'
>>> val.partition('*')[2]
'20'
>>> val = 's2-20'
>>> val.partition('*')[2]
''

It's smoother and performs surprisingly good - in fact a lot better than split:

>>> import timeit
>>> timeit.timeit('"s2*20".split("*")')
0.22846507105495942
>>> timeit.timeit('"s2-20".split("*")')
0.1685205617091654
>>> timeit.timeit('"s2*20".partition("*")')
0.1475598294296372
>>> timeit.timeit('"s2-20".partition("*")')
0.09512975462482132