This is not a complete answer, but is intended to expand on some of the comments above.

As mentioned by j.c. and Dylan Wilson in the comments, the set of unbased homotopy classes $[S^p\times S^q,X]$ is a quotient of the set of based homotopy classes $\langle S^p\times S^q,X\rangle$ under an action of $\pi_1(X)$. Since you say you are mostly interested in the case $X=BU(k)$, which is simply-connected, we may as well work with pointed homotopy classes.

The way I would try to analyse this (and perhaps this is the approach Ryan Budney had in mind in his comment) is using the cofibration sequence
$$
S^{p+q-1}\to S^p\vee S^q\to S^p\times S^q \to S^{p+q}\to S^{p+1}\vee S^{q+1}\to \cdots
$$
in which the first map is the attaching map of the top cell of $S^p\times S^q$. The fourth map is the suspension of this attaching map, therefore is null-homotopic. There results an exact sequence of pointed sets
$$
1 \to \langle S^{p+q},X\rangle \to \langle S^p\times S^q,X\rangle \to \langle S^p\vee S^q,X\rangle \to \langle S^{p+q-1},X\rangle
$$
which written in terms of homotopy groups becomes
$$
0\to \pi_{p+q}(X) \to \langle S^p\times S^q,X\rangle \to \pi_p(X)\oplus\pi_q(X)\to \pi_{p+q-1}(X).
$$
The last map is the Whitehead product. Hence if you know the homotopy groups of $X$ (which for $X=BU(k)$ are known in a stable range by Bott periodicity) and the Whitehead products, you have a good chance of describing the set $\langle S^p\times S^q, X\rangle$ (which by the way is not a group, since $S^p\times S^q$ is not a co-H-space).

**Edit:** In the comments, user DLIN asks about the case $X=Gl_N(\mathbb{C})$. We can say more in this case, since $X$ is a path-connected H-space. Therefore all Whitehead products in $\pi_*(X)$ are trivial, and the exact sequence of sets above becomes a short exact sequence of groups
$$
0\to \pi_{p+q}(X) \to \langle S^p\times S^q,X\rangle \to \pi_p(X)\oplus\pi_q(X)\to 0.
$$
Note that the middle term is now a group (using an H-space multiplication $\mu:X\times X\to X$). Furthermore, this sequence splits; the last map, which is given by restriction to the two sphere factors, is split by sending $(f,g)$ to $\mu\circ (f\times g)$.

Now suppose we know that $\langle S^p\times S^q, X\rangle$ is abelian. Then we could conclude that
$$
\langle S^p\times S^q, X\rangle \cong \pi_{p+q}(X)\oplus\pi_p(X)\oplus\pi_q(X).
$$
This would be the case, for example, if there are two commuting H-space structures on $X$ (see the Eckmann-Hilton argument).

somebasepoint of $X$ (trivially). Choosing a path from that one to $x$ gives a way of taking an arbitrary map and making it basepoint preserving. But there's ambiguity because there are different homotopy classes of paths between these two points. $\endgroup$ – Dylan Wilson Mar 23 '16 at 20:21