Let $V,W$ be two $k$-vector spaces of finite dimension. Let $G$ be a group and $H \leq G$ a subgroup, with $[G:H]=n$. Let $\{x_1,\dots x_n\}$ be a right trasversal for $G$ on $H$ and assume that $V$ is a $G$-module and that $W$ is an $H$-module.
I want to show that the two $G$-modules $W^G \otimes V$ and $(W \otimes V_H)^G$ are isomorphic, where $W^G$ is the extension of the "representation" $W$ to $G$, while $V_H$ is the restriction of the "representation" $V$ to $H$. The underlying vector space of $W^G$ is a direct sum of $n$-copies of $W$, while the underlying vector space of $(W \otimes V_H)^G$ is a direct sum of $n$ copies of $W \otimes V$. We know that there is a canonical isomorphism of $k$-vector spaces $f \colon(\bigoplus_{i=1}^n W) \otimes V \to \bigoplus_{i=1}^n W \otimes V$ which sends $(w_1,\dots,w_n) \otimes v \mapsto (w_1 \otimes v,\dots,w_n \otimes v)$.
I recall that the action of $G$ on $W^G$ (and so the one of $G$ on $W \otimes V_H$) is defined (using the embedding $G \to H^n \rtimes \mathrm{Sym}(n)$, s.t. $g \mapsto [(h(1,g),\dots,h(n,g)),\sigma(g)]$, where $x_ig=h(i,g)x_{i\cdot \sigma(g)}$) as $(w_1,\dots,w_n)^g=\big(w_{1\cdot \sigma^{-1}(g)}^{h(1\cdot\sigma^{-1}(g),g)},\dots,w_{n\cdot \sigma^{-1}(g)}^{h(n\cdot\sigma^{-1}(g),g)}\big)$. If i am not wrong, the map $f$ doesn't preserve the $G$-structure and so i don't know how to do to construct a $G$- isomorphism between those two vector spaces.
First of all, note that your notation is not really standard [as suggested in the comments, it is used, and so isn't "nonstandard" as I had originally written - what follows is still valid though] and may be confusing. Indeed, $W^G$ is often used to denote the fixed points of $W$ under $G$, where $W$ is a $G$-module, and $V_H$ the $H$-co-invariants of $V$ under $H$.
What you denote $W^G$ is usually denoted by $\mathrm{Ind}_H^GW$ : it's called the induction of $W$; it's also $k[G]\otimes_H W$.
Similarly, the thing you call $V_H$ is called the restriction of $V$ to $H$ and is usually denoted by $\mathrm{Res}_H^GV$.
Those are the notations I'll be using. These functors are intimately connected, in that they form an adjunction. If you know what that is, great, the rest will be much easier. If you don't, no problem, I'll explain what this tells us here, and why it's true.
Essentially it means that we have a universal property of $\mathrm{Ind}_H^GW$: first of all, for any $H$-module$W$, you have a specific $H$-map $\eta_W : W\to \mathrm{Ind}_H^GW$. Note that I should actually have written $\mathrm{Res}_H^G\mathrm{Ind}_H^GW$, because for this map $\mathrm{Ind}_H^GW$ is seen as only an $H$-module.
Now if you have an $H$-morphism $f:W\to \mathrm{Res}_H^GV$ (that is an $H$-morphism $W\to V$, where $V$ is actually a $G$-module), then there is a unique $G$-map $f^*:\mathrm{Ind}_H^GW\to V$ such that $f^*\circ \eta_W = f$.
This universal property characterizes $\mathrm{Ind}_H^GW$ uniquely up to $G$-isomorphism. Why is it satisfied ? Well with my presentation of the induction as $k[G]\otimes_HW$ it's pretty easy : $\eta_W$ simply sends $w$ to $1\otimes_H w$ and then if you have an $H$-map $f:W\to V$, where $V$ is a $G$-module, you can define $s:k[G]\times W\to V$, $(g,w) \mapsto g\cdot f(w)$, which is $H$-bilinear ($s(g,hw) = g\cdot f(hw) = ghf(w) = (gh)\cdot f(w) = s(gh,w)$), thus defines a unique $f^* : k[G]\otimes_H W\to V$, which is clearly $G$-linear; and it clearly satisfies $f^*\circ\eta_W = f$, and it's clearly the unique one to do so.
Now we have a strategy to attack the proof of your isomorphism : show that the LHS satisfies the universal property of the RHS. A very succint way of writing this is by using the natural isomorphisms given by the adjunction. I'll begin by doing this, to show the efficiency of the categorical formalism, but then I'll actually dive into what maps we are using and why they satisfy such and such properties: that'll be useful only if you don't know the categorical formalism.
Here goes : $$\hom_G(\mathrm{Ind}_H^G(W\otimes \mathrm{Res}_H^G V), -) \cong \hom_H(W\otimes \mathrm{Res}_H^GV, \mathrm{Res}_H^G-) \cong \hom_H(W,\hom_k(\mathrm{Res}_H^GV,\mathrm{Res}_H^G-)) \cong \hom_H(W,\mathrm{Res}_H^G\hom_k(V,-)) \cong \hom_G(\mathrm{Ind}_H^GW, \hom_k(V,-))\cong \hom_G(\mathrm{Ind}_H^GW\otimes V, -)$$
all these isomorphisms are natural (note that the $G$-action on $\hom_k(V,-)$ is the usual one, i.e. $(g\cdot f) (v) = gf(g^{-1}v)$), thus by the Yoneda lemma, we get $\mathrm{Ind}_H^G(W\otimes \mathrm{Res}_H^G V)\simeq \mathrm{Ind}_H^GW\otimes V$ (naturally in $V,W$).
Ok, now let's see what this means less abstractly ! Let's check that the LHS has the universal property of the RHS, by hand.
To do this, we must begin by finding an $H$-map $W\otimes V\to \mathrm{Ind}_H^GW\otimes V$ (where I abused notation by not keeping track of the restrictions, I hope it's clear what it means - if not I can edit my answer).
But this is easy : I have a map $\eta_W : W\to \mathrm{Ind}_H^GW$, and the identity $V\to V$ (which is clearly an $H$-map !), so I can tensor them and get $\eta : W\otimes V\to \mathrm{Ind}_H^GW\otimes V$, which is an $H$-map. It now remains to check that it satisfies the universal property above.
So let $M$ be a $G$-module, and suppose we have an $H$-map $f:W\otimes V\to M$. Now this corresponds to a map $f^c: W\to \hom_k(V,M)$, by universal property of the tensor product. The fact that $f$ was $H$-linear tells us something about $f^c$. Indeed, $f^c(w)(v) = f(w\otimes v)$. So $f$ being $H$-linear means (with obvious notation and quantification) $f(h\cdot (w\otimes v)) = hf(w\otimes v) = hf^c(w)(v)$, but on the other hand $f(h\cdot (w\otimes v)) = f(hw \otimes hv) = f^c(hw)(hv)$. Now if we want to see what this means for $f^c(hw)$, we have to compute $f^c(hw)(v)$. But this is $f^c(hw)(hh^{-1}v) = hf^c(w)(h^{-1}v)$. But that's marvelous ! Indeed, what we just proved was that $f^c(hw) = h\cdot f^c (w)$, with the usual action on $\hom_k(V,M)$ given above ! So $f^c$ is $H$-linear.
But wait a second, the usual action of $H$ on $\hom_k(V,M)$ is just the restriction of the one of $G$, because both $V,M$ are $G$-modules, so this tells us that (by the universal property) we get a map $f^§ :\mathrm{Ind}_H^GW\to \hom_k(V,M)$, which is $G$ linear, and such that $f^§\circ \eta_W = f^c$. Moreover, it's unique with respect to those requirements !
So now by universal property of the tensor product, we get a map $f^* : \mathrm{Ind}_H^GW\otimes V\to M$. The $G$-linearity of $f^§$ can, as above, be used to show that $f^*$ is itself $G$-linear. Moreover, $f^*(\eta(w\otimes v)) = f^*(\eta_W(w)\otimes v) = f^§(\eta_W(w))(v) = f^c(w)(v) = f(w\otimes v)$ : $f^*\circ \eta = f$ !
Ok and now if $g$ is another $G$-map satisfying $g\circ \eta = f$, the above computation shows that $G$ induces a $G$-map $g' : \mathrm{Ind}_H^GW\to \hom_k(V,M)$ such that $g'\circ \eta_W = f^c$, so that $g' = f^§$, and so $g=f^*$, whence the uniqueness.
What can we conclude from this ? Well the LHS satisfies the universal property of the RHS, so there is a unique isomorphism $\mathrm{Ind}_H^G(W\otimes \mathrm{Res}_H^G) \to \mathrm{Ind}_H^GW\otimes V$ that commutes with the $\eta$'s !
Ok this answer is already way too long but technically I'm not done yet, because I would still have to show that my presentation of the induction is the same as yours. I won't prove it here, and leave it to you as an exercise, with the following hint : $$k[G]\otimes_H W \simeq (\bigoplus_{\sigma \in G/H} k\sigma H) \otimes_H W \simeq \bigoplus_{\sigma \in G/H}(k\sigma H\otimes_H W) = \bigoplus_{\sigma \in G/H}(k\sigma \otimes_H W) \simeq \bigoplus_{\sigma \in G/H}\sigma W$$, so it just remains to ellucidate how $G$ acts on that.