Given that he should not be able to guess hash chains he has no preimage for with probability $> 2^{-n}$, it does not seem to threaten the security of the overall signature at first glance, is that correct?
No, that is not correct.
The attacker can forge if he can find a preimage where he knows the image that appears in all $x$ of the chains that preimage implies. This analysis will assume that he has a valid signature to some message, and possibly some additional information.
Because of how LMS and XMSS implement their chains (using some of the bits directly as a chain location ("the message bits") and using an 'inverse checksum' to compute the chain location ("the checksum bits"), then any modification to the message will cause a location in some chain to decrement; to find that that, that requires computing a hash preimage. Hence, the attacker is limited to 'signing' the original message (and because that message really is the hash of the actual signed message, that isn't any easier).
As an aside (because this turns out to be important in the analysis) the attacker cannot directly control the "message" being signed by the Winternitz scheme - instead, he specifies a message and that is hashed, and then that message is signed with Winternitz. What that means is that the attacker cannot directly control the message bits; instead, he ends up having to hash messages and hope he gets one with the right message and checksum digits.
With that in mind, if you give the attacker some of the earlier chain locations, then he has more flexibility. He can now consider messages that decrement that particular chain; as long as the chains he doesn't have information stay the same or increase, he can forge.
Let us take a simple example; suppose he knows the immediate preimage to one of the message chains; what this means (assuming the lsdigit of the checksum wasn't zero - that causes complications in the analysis), then he can forge if he can find a message that hashes to a value that is either exactly the hash of the valid message that was signed (a second preimage) or decreases that one digit by one and increases some other digit (either another message digit or the lsdigit of the checksum) by one. Because there are approximately $x-1$ such other message digits (ignoring the upper checksum digits), that means that the attacker has about $x$ potential hash targets, rather than one. That means that with this minimal leakage, the security is about $n - \log_2 x$ bits, rather than $n$ bits. This may not seem like much; however the leakage was quite minimal ($y=1$ in your terminology, and you didn't start if the attacker learned the immediate preimage, the base of the chain or somewhere in the middle).
The analysis gets rather more complex (and the security loss is greater) if the leakage is larger (or if the leakage is in the checksum digits; that gets messy).
