The security level of a hash function is determined by its output size. In general, a hash function is considered cryptographically secure when it is collision resistant and provides security level $b = n/2$. Hence, it is not wrong to describe Lamport's scheme that way. However, the description probably was done that way to abstract away some details:
If you want to get $b$ bits security and do a classical message digest, you will need a hash function with $m = 2b$ bit outputs for the digest. However, for the "internal" hash function used to map secret key elements to public key elements you are fine with $n = b$ bit outputs as you just need one-wayness for this function. Now, if you change the message digest to a randomized message digest using $h = H(R,M)$ for randomness $R$ and message $M$, you can get away with $m = n = b$ bit output length for the message digest (if you do everything correctly). This is more about the message digest than about the hash function used inside of Lamport and hence distracting from the actual topic: How does Lamport's scheme work.