repeating-key xor and hamming distance
Yes, you are remembering correctly.  Yes, this is a reasonable method to find the key length.

The reason why this works is because, typically, the plaintext is not uniformly random.  For instance, rather than a random bit-string, the plaintext might be some English text, encoded in ASCII.  If $X,Y$ represent two random English letters, encoded in ASCII, then the expected value of the Hamming distance $\text{wt}(X \oplus Y)$ is maybe 2-3 bits.  In contrast, if $U,V$ are two random 8-bit bytes, then the expected value of the Hamming distance $\text{wt}(U \oplus V)$ is 4 bits, significantly larger.  If you look at sequences of multiple characters, rather than a single letter at a time, the difference becomes even larger.

How does this apply to your situation?


Well, if you have correctly guessed the key length, then your ciphertext consists of $X\oplus K$ and $Y\oplus K$ (as Dilip Sarwate explains), where $X,Y$ come from the plaintext distribution.  Now notice that the Hamming distance between these two is the same as the Hamming distance between $X$ and $Y$, namely, it is $\text{wt}(X \oplus Y)$.  As we explained before, you can expect this might be maybe 2-3 bits times the length of $X$ measured in bytes.
In contrast, if you guessed the key length incorrectly, then you're looking at ciphertexts of the form $X \oplus K$ and $Y \oplus K'$.  The Hamming distance between the two basically boils down to the Hamming distance between $U$ and $V$, where $U$ and $V$ are uniformly randomly distributed (since $K,K'$ are uniformly randomly distributed), and thus is $\text{wt}(U \oplus V)$.  As explained before, you can expect this should be approximately 4 bits times the length of $X$ measured in bytes.


So, as you can see, the Hamming distance is significantly less when you've guessed the key length correctly.

For a vaguely similar method, read about the Index of coincidence; you can expect it to be more effective in some cases, and less effective in others.