This is really a comment which got too long.

Personally, I always find this one rather confusing. If you think in terms of modules over group rings, we want to show that $U \otimes (\mathbb{C}[G]\otimes_{\mathbb{C}[H]} W) \cong \mathbb{C}[G] \otimes_{\mathbb{C}[H]} (U\otimes W)$. The $G$-equivariant isomorphism is not given by sending $u\otimes (x\otimes w)$ to $x \otimes (u\otimes w)$. There's a lot wrong with this formula, but that's not the point.

The point is that to get the right formula, one really needs to remember exactly how the universal property of induction works. I don't have Fulton and Harris in front of me to see what they say, but Serre's book has a good discussion of induction which will lead one right to the answer.

Also, unless I'm confused, this really seems to depend on the structure of $\mathbb{C}[G]$ as both a ring and as a $\mathbb{C}[H]$-module. One needs to know that it's a free $\mathbb{C}[H]$-module, and that it has a decomposition as a $C[H]$-module into summands isomorphic to $C[H]$ that are permuted by the units of the ring $C[G]$. One could ask, for morphisms of rings $C\to R\to S$, when it's true that for an R-module M and an S-module N we have the formula $N \otimes_C (S\otimes_R M) \cong S \otimes_R (N \otimes_C M)$ (as S-modules). (Above I wrote $\otimes$ instead of $\otimes_{\mathbb{C}}$; now C is the ground ring). I don't know how to prove this without assuming S has the sort of structure mentioned above (free as an R-module, etc.).