For a given pair of cipher text $(c, c')$, you can compute $x = c \oplus c'$ which is the XOR of the two ciphertexts and is also equal, as you note, to the XOR of the two corresponding plaintexts. Then, for any pair of potential messages $(m, m')$ such that $m \oplus m' = x$, you can compute a corresponding key $K = m \oplus c$; you can verify that $Enc(K, m, m') = (c, c')$. Therefore, the information you have (the ciphertexts $c$ and $c'$) gives you the XOR of the two plaintexts, but since any pair of messages which matches that information is still a possibility, you have no more information than that.
An other way to see it: given a One-Time Pad encryption of a message $m$ with a key $K$ (you know $c = m \oplus K$, but not $m$ or $K$), you can create a random $x$ and compute $c' = c \oplus x$. You then know that $Enc(K, m, m') = (c, c')$ and $m \oplus m' = x$. If you can extract out of that more information than the XOR of $m$ and $m'$ (which the value $x$ you chose), then, congratulations, you have broken One-Time Pad. Given the proven impossibility thereof, one must conclude that out of $(c, c')$ you learn only $m \oplus m'$ and no more.
Mind you, for messages $m$ and $m'$ which are not completely random (e.g. plaintext messages which make sense), the XOR of $m$ and $m'$ is a lot of information; this is the infamous "Two-Times Pad" which has lead to actual decryption in some historical cases (the Wikipedia page on OTP gives a few example, with the Soviet Union in the role of the loser).