Let $f,g$ be functions in $C^A$ and $C^B$ respectively.
Let $\boxtimes:C^A \times C^B \to (C\times C)^{A \times B}$ s.t.
$f\boxtimes g(a,b)=(f(a),g(b))$
It seems not the tensor product, nor Cartesian product. Then can we call it direct product? But it seems the term 'direct product' often used on operator between structures.
On
Your map is some transpose of the Cartesian product.
Given a function $f : A \to B$ and another function $g:C \to D$, their Cartesian product is
$$ f \times g : A \times C \to B \times D : (a,c) \mapsto (f(a), g(c)) $$
The notion of "transpose" enters the picture in the sense that you are switching from thinking about a function $A \to C$ to thinking of an element of $C^A$.
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.