Hash collision resistance requirements for Lamport signatures
Yes, it makes sense to truncate the hash to 128 bits. The security proof actually says that if finding a preimage for F requires effort $2^{n}$, then breaking the Lamport signature scheme with $G$ having k-bit digests requires effort $2^{n}/2k$. So strictly speaking, with $F$ truncated to 128 bits and $G$ having 256 bits $(2k=512=2^{9})$, you will have $128-9=119$ bits of security.

This is not an artifact of the security proof: since there are $256*2=512$ public keys, an attacker has 512x more chances to find a preimage compared to a normal, single preimage attack. 

In the Merkle signature scheme, the result is the same but you would multiply k by the number of messages that can be signed. EG for 1 million $(2^{20})$ messages, the security level is $128-9-20=99$ bits.

The wikipedia article is about $F$ only: the "input" is one of the private keys, the "output" is the corresponding public key and "message digest size" is the output size of $F$. I think that it just attempts to say that when the private key is very short and regardless of the output size of $F$, it is trivial to find a preimage.