Sadly, no. In the walkthrough at around 4:30 Zamyla advises "advance to the next letter in keyword only when plaintext is a letter". It is conditional. Conditional suggests an
if statement. There is no loop (ie no
k = strlen(s) % strlen(key) looks like an attempt to handle the wraparound. Be warned, the index of key (assuming
k here) is independent of the plaintext. It only depends on (the length of) the key. Every time you use a letter from the key, increase the index (
k) by 1. But never let it be greater than the length of the key (that's where modulo comes in). Review the countoff example in the walkthrough at 6:05.
There are other details that will give you trouble; you are going to run into problems with the way you are handling case. Case of the plaintext and case of the key are independent of each other, great hint in walkthrough at 0:14. Think about how you could convert
key[k] to a "caesar number" before you encipher the plaintext letter. Then you need only consider the case of the plaintext.
Suggest you rewatch the walkthrough, write the pseudocode as suggested, then attack the problem step-by-step. If you got caesar you can definitely get this.