The formal definition of a standard Turing machine $M$ is as follows (from An Introduction to Formal Languages and Automata, 6e - Peter Linz):
$$M = (Q, \Sigma, \Gamma, \delta, q_0, □, F)$$
Aside from the other elements, $F$ is the symbol that denotes the finite set of accepting (a.k.a. final) states, and $F \subseteq Q$.
There is no restriction that states how many accepting states a particular Turing machine must have. The machines will differ from problem to problem. Nor is there any statement that states having a single accepting state is the best, and that other accepting states are "unnecessary."
I'm not sure if this is comparable to multi-track Turing machines vs. standard Turing machines, though. As you said, multi-track Turing machines are usually for the sake of simplicity and are equivalent to their standard counterparts, whereas the number of final states is usually a necessity depending on the language. The two seem to be fundamentally different issues to me.