Is the verifier for an $NP$ and its $co-NP$ the same?
I think you might be slightly confusing your definition for co$NP$. To show that a certain graph doesn't contain a Hamiltonian path, you need to show that all paths in it aren't Hamiltonian, and providing a specific path as a "counter example" isn't enough. 

In general, $coNP$ isn't defined using a witness the same way $NP$ is, because if it did than it would be the same as $NP$, and for now this is an open problem (we don't know if the two are equal, and they are usually believed to be different). Therefore, it doesn't really make sense to talk about a witness for a $coNP$ problem the same way you talk about a witness for an $NP$ problem.

Specifically, for a certain language in $coNP$ and a witness-based TM $M$ accepting it, $x$ is in the language iff $M$ accepts $(x,v)$ for all witnesses $v$. This is in contrast to a TM for a language in $NP$ that needs to accept only for a single witness.

I would like to add that for me, studying about the Polynomial Hierarchy using its witness-related definitions helped me understand $coNP$ better, but it isn't a necessity.