You can always reduce anything to anything, but in this case, the reduction is relevant because it only adds a polynomial factor. Showing that a problem can be reduced to another one in polynomial time is core to proving np-completeness.

That means that tsp and tso-opt are in the same ballpark of complexity. The reason tsp-opt isn't called np-complete isn't related to its complexity, but to the fact that only decision problems get to be called that.

As I said, there are problems where the decision problem and the optimization problem are completely unrelated in terms of complexity, but tsp isn't one of those.

Nope, the usual way of reducing optimization problem is to ask the corresponding yes/no question - for some fixed value x, is the optimal value greater than x?

If you can solve the decision problem, you can solve the optimization problem by doing a binary search on the x.