In a Vigenère cipher, the $i$-th ciphertext letter $c_i$ is calculated as $p_i + k_{i \bmod \ell}$, where $p_i$ is the $i$-th plaintext letter, $k_i$ is the $i$-th key letter, and $\ell$ is the length of the key. Thus, the difference between two ciphertext letters at positions $i$ and $j$ is $$c_i - c_j = p_i - p_j + c_{i \bmod \ell} - c_{j \bmod \ell}.$$
Normally, for this difference to be zero, the difference between the key letters would have to just happen to cancel out the difference between the plaintext letters. When $i \equiv j \pmod \ell$, however, the last two terms cancel out, leaving just $p_i - p_j$, which, since the plaintext typically has an uneven letter frequency distribution, is more likely to be zero than one would otherwise expect.
Now, consider the autokey cipher, where (beyond the first $\ell$ letters) $c_i = p_i + p_{i-\ell}$. Now the difference between two ciphertext letters is given by $$c_i - c_j = p_i - p_j + p_{i-\ell} - p_{j-\ell}.$$ This time, when $i - j = \ell$, it's the middle terms that cancel, leaving just $p_i - p_{j-\ell}$ $=$ $p_i - p_{i-2\ell}$.
Again, it's a lot more likely for a single pair of plaintext letters to be identical than it is for the differences of two pairs of letters to cancel out just by chance. Exactly how much more likely depends on details of the plaintext frequency distribution, but the qualitative behavior is quite robust and shows up for most naturally occurring letter frequencies.
Ps. Note that the conditions for this cancellation to occur are different. For the Vigenère cipher, cancellation occurs whenever $i - j$ is a multiple of the key length $\ell$, producing a characteristic comb-shaped autocorrelation plot. For an autokey cipher, however, cancellation only occurs when $i - j$ exactly equals the key length, resulting in just a single peak.