On the other hand, if you wish to allow only homotopies $f_t$ and $g_t$ such that $f_t=g_t$ on $S^2$ for all $t$, then such pairs $(f,g)$ are equivalent to maps $h:S^3\to X$. Homotopy classes of maps $h:S^3\to X$ without basepoint conditions are in one-to-one correspondence with orbits of the action of $\pi_1(X)$ on $\pi_3(X)$, and this set of orbits can be nontrivial.

If you do not require $f=g$ on $S^2$ during homotopies, then the problem becomes trivial since every map $B^3\to X$ is homotopic to a constant map. Since we may assume $X$ is path-connected, this means $f$ and $g$ are homotopic.

Interesting use of language here (twice) that I don't recall seeing before: "If [X], if [Y], then [Z]." Is this equivalent to "If [X] and [Y], then [Z]"? Or maybe "If [X] such that [Y], then [Z]".

Also, $T^*$ is a cellulation (a decomposition into cells), not a triangulation as stated in the question.

You are talking about a product that involves just homology and not cohomology, so this is not the cap product. Instead it is usually called the intersection product. A classical reference for the intersection product is the textbook by Seifert and Threlfall. A more recent textbook source is Bredon's "Topology and Geometry". Bredon attributes the intersection product to Lefschetz. The intersection product is a perfectly well-behaved product, with the right hypotheses, so it's not clear why you use the words "evil", "fails", and "ill-defined" in reference to it.