Try using Frobenius reciprocity. Let $V$ and $W$ be two representations of $H$, and let $U$ be a representation of $G$. Consider first the space:
$$Hom_G \left(U, Ind_H^G (V \otimes W) \right) \cong Hom_H \left( U, V \otimes W \right),$$
by Frobenius reciprocity.

On the other hand, one can consider the space:
$$Hom_G(U, Ind_H^G V \otimes Ind_H^G W).$$
This is canonically isomorphic to
$$Hom_G(U \otimes Ind_H^G V', Ind_H^G W),$$
where $V'$ denotes the dual representation of $V$. By Frobenius reciprocity again, this is isomorphic to:
$$Hom_H(U \otimes Ind_H^G V', W).$$
This is canonically isomorphic to
$$Hom_H(U, (Ind_H^G V) \otimes W).$$
Now, we are led to compare the two spaces:
$$Hom_H(U, V \otimes W), \quad Hom_H \left( U, (Ind_H^G V) \otimes W \right).$$

There is a natural embedding of $V$ into $Res_H^G Ind_H^G V$. This gives a natural map:
$$\iota: Hom_H(U, V \otimes W) \rightarrow Hom_H \left( U, (Ind_H^G V) \otimes W \right).$$

Using complete reducibility, let us (noncanonically) decompose $H$-representations:
$$Res_H^G Ind_H^G V \cong V \oplus V^\perp.$$
It follows that
$$Hom_H \left( U, (Ind_H^G V) \otimes W \right) \cong Hom_H \left(U, V \otimes W \right) \oplus Hom_H \left( U, V^\perp \otimes W \right).$$

It follows that $\iota$ is injective. This explains (via Yoneda, if you like) why $Ind_H^G(V \otimes W)$ is canonically a subrepresentation of $Ind_H^G V \otimes Ind_H^G W$. It also explains that computation of "the rest" of $Ind_H^G V \otimes Ind_H^G W$ -- the full decomposition into irreducibles -- requires Mackey theory: the decomposition of $Res_H^G Ind_H^G V$. There can be no neat answer, without performing this kind of Mackey theory.

`$[G:H]$`

, so your second module is typically bigger than your first. – Jim Humphreys Mar 15 '10 at 20:10