I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V_i$ from $V$. This is a subtle and oft-overlooked point in representation theory, in my opinion. Many texts prove that an irrep of $G_1 \times G_2$ is isomorphic to a tensor product of an irrep of $G_1$ with an irrep of $G_2$. The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes.

Here's a categorical explanation of the theorem: Let $G_1$ and $G_2$ be finite groups, and let $\pi$ be an irrep of $G_1 \times G_2$ on a complex vector space $V$.

Then, for every pair $(\rho_1, W_1)$, $(\rho_2, W_2)$ of representations of $G_1, G_2$, one gets a complex vector space:
$$H_\pi(\rho_1, \rho_2) := Hom_G(\rho_1 \boxtimes \rho_2, \pi).$$
In fact, this extends to a contravariant functor:
$$H_\pi: Rep_{G_1} \times Rep_{G_2} \rightarrow Vec.$$
Here we use categories of finite-dimensional complex representations and vector spaces.

This is also functorial in $\pi$, yielding a functor:
$$H: Rep_G \rightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ],$$
where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$). What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep_{G_1}$ and $Rep_{G_2}$. By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories:
$$Rep_{G_1} \times Rep_{G_2} \hookrightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ].$$

It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep_G$, the functor $H_\pi \in [Rep_{G_1} \times Rep_{G_2}, Vec]$ is representable. It is not uniquely representable, but it is uniquely representable up to natural isomorphism.

Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi_1 \boxtimes \pi_2$ for some irreps $\pi_1, \pi_2$ of $G_1, G_2$, respectively. The pair $(\pi_1, \pi_2)$ is not unique, but the triple $(\iota, \pi_1, \pi_2)$ is unique up to unique isomorphism. This is usually good enough.