Let's define a new problem, TSP-cost-optimize, which returns the cost of the best Hamiltonian. It's NP-Complete, because it reduces to TSP-decide: Just do a binary search on the space of possible scores; even if that space is exponential, we're fine, since binary search runs in the log of the size of the space.
TSP-cost-optimize's certificate can be the steps of the binary search; that will be polynomial in size, since it's a polynomial number of TSP-decide certificates, and you can verify it in polynomial time.
Given TSP-cost-optimize, I can in fact provide a polynomial-time-verifiable certificate for the result of TSP-optimize: The result of TSP-cost-optimize on the same graph! So under your definition of NP, TSP-optimize is in fact in NP. (This also makes sense if you use the non-deterministic Turing machine definition of NP.)
What's not clear to me is whether given the result of TSP-cost-optimize, you can actually find a shortest path in polynomial time. But I expect you probably can.