There are 2 main problems with your code that cause the algorithm to output the wrong answer.

if i == 0 or j == 0 in line 16

Just following the video shows that this line makes no sense when s1[1] != s2[j], because the longest common subsequence of "ab" and "a" has length 1 although your algorithm sets matrix[0][1] = 0 for this example. So you need to remove this if statement. While your at it you have to consider what max(matrix[i-1][j], matrix[i][j-1]) does for i == 0 or j == 0. Now there are two different approaches:

1. The explicit one:

   max(matrix[i-1][j] if i != 0 else 0, 
       matrix[i][j-1] if j != 0 else 0)
   

2. The implicit one:

   max(matrix[i-1][j], matrix[i][j-1])
   

   This one works, because in Python negative indices are used to get the last item of a list and these items are 0 in this case.

cs += s1[i] in line 11/14

For example if you found that the longest common subsequence of "a" and "abcd" is "a", your algorithm sets the longest common subsequence for "a" and "abcda" as "aa", which doesn't make sense. I am struggling to explain why it does not work like that, so i suggest you look at a few examples, maybe using http://pythontutor.com/visualize.html

Solution

To approach both problems you can use the matrix to store the longest common subsequence you found for the smaller problems. You end up with this:

def lcs(s1, s2):
    matrix = [["" for x in range(len(s2))] for x in range(len(s1))]
    for i in range(len(s1)):
        for j in range(len(s2)):
            if s1[i] == s2[j]:
                if i == 0 or j == 0:
                    matrix[i][j] = s1[i]
                else:
                    matrix[i][j] = matrix[i-1][j-1] + s1[i]
            else:
                matrix[i][j] = max(matrix[i-1][j], matrix[i][j-1], key=len)

    cs = matrix[-1][-1]

    return len(cs), cs

print(lcs("abcdaf", "acbcf"))  

This specific implementation only returns one possible result. You can try to implement an algorithm that gives all of the longest common sequences as an exercise. Maybe have a look at the Wikipedia page as suggested by גלעד ברקן

How long does it take to "get" why your code does not work?

There is obviously no clear answer. It always helps to think about examples and in the case of algorithms Wikipedia often has a good pseudocode, which you can base your implementations on. When you are familiar with the concepts and datastructures involved in the algorithm, you should be able to implement it within a day, I would say (but I am definitely no expert). In general searching for logical bugs in your code, can take multiple days, depending on the size of the code. To practice this kind of structured, algorithmic and mathematical thinking I can highly recommend projecteuler.net.

For those who look for a built-in solution:

from difflib import SequenceMatcher

str_a = "xBCDxFGxxxKLMx"
str_b = "aBCDeFGhijKLMn"
s = SequenceMatcher(None, str_a, str_b)

lcs = ''.join([str_a[block.a:(block.a + block.size)] for block in s.get_matching_blocks()])
# lcs = 'BCDFGKLM'

By just getting the length of LCS using following code segment, you will find the max length is 14, so BurningKarl's solution works for me.

def longestCommonSequence(s1, s2, s1Index, s2Index, arr):
    if s1Index ==-1 or s2Index == -1:
        return 0
    if(arr[s1Index][s2Index] != None):
        return arr[s1Index][s2Index]

    if s1[s1Index] == s2 [s2Index]:
        result = 1+ longestCommonSequence(s1, s2, s1Index -1, s2Index -1, arr)
    else:
        result= max(longestCommonSequence(s1, s2, s1Index -1, s2Index, arr), longestCommonSequence(s1, s2, s1Index, s2Index -1, arr))
    arr[s1Index][s2Index] = result
    return result   

s1="AAACCGTGAGTTATTCGTTCTAGAA" 
s2="CACCCCTAAGGTACCTTTGGTTC" 

a = [[None for i in range(len(s2))] for j in range(len(s1))]
print(longestCommonSequence(s1, s2, len(s1)-1, len(s2)-1, a))