For the viewpoint of category theory, your map is just $f\times g$ -- it is the image of $f$ and $g$ under the product bifunctor $(-)\times(-)$.
More verbosely, if you compose $f$ and $g$ with the projection maps from $A\times B$, then you get maps $f\circ \pi_1: A\times B \to C$ and $g\circ \pi_2: A\times B \to C$, which factor through $C\times C$ by the universal property of the latter. The mediating morphism is excatly $f\times g$.
If you consider $\mathbf{Set}$ to be a monoidal category by declaring $\times$ to be $\otimes$, then $f\times g$ is indeed the tensor product $f\otimes g$.
On the other hand, in ordinary set theory, we usually identify a function with its graph: $f=\{\langle x,f(x)\rangle\mid f(x)\text{ is defined}\},$ and in that sense your $f\boxtimes g$ is of course not the cartesian product of $f$ and $g$. It is closely related though: $ f\boxtimes g = \{ \langle\langle a,b\rangle,\langle c,d\rangle\rangle \mid \langle\langle a,c\rangle,\langle b,d\rangle\rangle \in f\times g\}$ which could be seen as a more vivid justification for Hurkyl's "transpose" terminology.