Why is Lamport-Diffie secure?
Lets assume an adversary knows one LD signature for message $M$ as well as public key $pk$ and can generate a forgery for an arbitrary different message $M'$. Clearly, there exists at least one bit position where the messages differ, i.e. $M_i \neq M'_i$, as $M \neq M'$. So, for simplicity assume there is only one bit difference. Then the adversary can take all signature values for the signature on $M'$ from the signature on $M$ besides the value for this single bit position. For the signature value of this bit it must hold that $f(sig_i) = pk_{i,M'_i}$ as otherwise verification would fail.  

So if the adversary is able to generate a valid forgery it obviously is able to find a value $x$ that satisfies $f(x) = pk_{i,M'_i}$ for given $pk_{i,M'_i}$. This is per definition finding a preimage of $f$. 

To get back to your question: The adversary might learn some information about $sk_{i,M'_i}$ from $pk_{i,M'_i}$. But that is not enough to produce a forgery as a forgery requires knowledge of a complete value $x$. Even more: it is not required that $x = sk_{i,M'_i}$ for the forgery to be valid.