I have just started looking into the cracking of the Vernam cipher with two ciphertexts encrypted with the same key by XORing them and then crib dragging and I was wondering if it is made easier if you have more ciphertexts like 3 or 4, and if so how that would be used to the crackers advantage?
Yes, in practice the more ciphertexts you have with the same key, the easier it is to break the encryption:
With more ciphertexts, you'll have more chances of hitting a crib, the more plaintext you'll reveal when you do, and the easier it is to be sure that a particular match is correct and not just plausible-looking by chance.
The more ciphertexts you have, the more opportunities you'll have for extending a decrypted segment, since you only need to guess how one of the messages continues outside the segment.
This homework exercise with 11 messages XORed with the same key is a good example: you can solve the puzzle almost entirely simply by starting with the assumption that the plaintexts (mostly) consist of ASCII letters and spaces, and observing that — due to the way the ASCII code is arranged — XORing a space with a letter has a specific and distinctive effect (it just flips the case of the letter) that is easily distinguishable from the result of XORing two letters (producing numbers, punctuation and/or non-printable control characters) or two spaces (which just produces a null byte). With 11 plain English ASCII messages, most positions in the aligned ciphertexts end up containing a space in at least one of the messages, and the few remaining gaps are then easy to fill in.