Here is an other proof for the case $H$ trivial.
Definition: Let $\langle U,V \rangle$ be the direct sum of all the irreducible representations generated by the irreducible complex representations $U$ and $V$ of the finite group $G$, for $\otimes$.
Lemma: $G_{(U \otimes V)} \supset G_{(U)} \cap G_{(V)} = G_{(\langle U,V \rangle)}$.
proof: The first inclusion is clear. Next, if $G_{(U)} \cap G_{(V)} \subset G_{(W_i)}$, then $G_{(U)} \cap G_{(V)} \subset G_{(W_1)} \cap G_{(W_2)}$ $ \subset G_{(W_1 \otimes W_2)}$, so $G_{(U)} \cap G_{(V)} \subset G_{(\langle U,V \rangle)}$. Now, $U,V \le \langle U,V \rangle$, so $G_{(\langle U,V \rangle)} \subset G_{(U)}, G_{(V)}$. $\square$
Next by Frobenius reciprocity, if $W \le U \otimes V$ then $U \le W \otimes \overline{V}$ and $V \le \overline{U} \otimes W$.
So $G_{(W \otimes \overline{V})} \subset G_{(U)}$ and $G_{(\overline{U} \otimes W)} \subset G_{(V)}$. Finally, $\langle W,\overline{V} \rangle = \langle W,V \rangle$ and $\langle \overline{U}, W \rangle = \langle U, W \rangle$ (because $G$ is finite), so by the previous lemma: $G_{(U)} \cap G_{(W)}$ and $G_{(W)} \cap G_{(V)} \subset G_{(U)} \cap G_{(V)}$. $\square$
Remark on the general case
If $G_{(U^H)} = G$ and $G_{(V^H)} \subsetneq G$ and if $ W \le U \otimes V$ then $G_{(W^H)} = G$.
So in this case $G_{(U)} \cap G_{(W)} \not\subset G_{(U)} \cap G_{(V)}$. Example: $G = S_4$, $H=G_1 = S_3$, $U$ the non-trivial $1$-dim. irr. rep. and $V$ the $3$-dim. irr. rep. with $V^H$ is $1$-dim.
Next, if $G_{(U^H)}$, $G_{(V^H)} \subsetneq G$, then for $[G:H] \le 30$ and $\vert G \vert \le 10^3$, every $W \le U \otimes V$ is working.
Conclusion, the proof above "could" be generlalized by assuming $H$ core-free and $G_{(U^H)}$, $G_{(V^H)} \subsetneq G$.