Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).

We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. Now, we can see the representation $\mathrm{Res}\mathrm{Ind}V$ as the left $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$. Then, there is a canonical $H$-equivariant injection $V\to \mathrm{Res}\mathrm{Ind}V$ given by $v\mapsto 1\otimes v$, and there is a canonical $H$-equivariant projection $\mathrm{Res}\mathrm{Ind}V\to V$ given by $g\otimes v\mapsto gv$ for $g\in H$ and $g\otimes v\mapsto 0$ for $g\not\in H$. This projection splits the injection, and therefore the representation $V$ is canonically a direct summand of the representation $\mathrm{Res}\mathrm{Ind}V$. Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.

"Canonically" means "canonically with respect to $U$ and $V$ and kind-of canonically with respect to $G$ and $H$" here. "Kind-of canonically with respect to $G$ and $H$" means that it's functorial with respect to maps which preserve both "lying in $H$" and "not lying in $H$", and I think we can't do better. As opposed to Robin's and Marty's proof, we don't need to rely on some randomly chosen system of representatives of cosets or double cosets.

Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).

We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. Now, we can see the representation $\mathrm{Res}\mathrm{Ind}V$ as the left $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$. Then, there is a canonical $H$-equivariant injection $V\to \mathrm{Res}\mathrm{Ind}V$ given by $v\mapsto 1\otimes v$, and there is a canonical $H$-equivariant projection $\mathrm{Res}\mathrm{Ind}V\to V$ given by $g\otimes v\mapsto gv$ for $g\in H$ and $g\otimes v\mapsto 0$ for $g\not\in H$. This projection splits the injection, and therefore the representation $V$ is canonically a direct summand of the representation $\mathrm{Res}\mathrm{Ind}V$. Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.

"Canonically" means "canonically with respect to $U$ and $V$ and kind-of canonically with respect to $G$ and $H$" here. "Kind-of canonically with respect to $G$ and $H$" means that it's functorial with respect to maps which preserve both "lying in $H$" and "not lying in $H$", and I think we can't do better. As opposed to Robin's and Marty's proof, we don't need to rely on some randomly chosen system of representatives of cosets or double cosets.

Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).

We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. But Mackey's double coset theorem says that $\mathrm{Res}\mathrm{Ind}V\cong \bigoplus\limits_{s\in H\backslash G/H}V_s$, so that in particular $V$ is canonically a direct summand of $\mathrm{Res}\mathrm{Ind}V$ (because $V=V_s$ for $s$ being a representative of the double coset of $1$). Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.

"Canonically" means "canonically with respect to $U$ and $V$". Canonicity with respect to $G$ and $H$ doesn't seem to follow from this, but at least the injection $\mathrm{Ind}\left(U\otimes V\right)\to \mathrm{Ind}U\otimes\mathrm{Ind}V$ can be made canonical with respect to $G$ and $H$ (because there is a canonical injection $V\to \mathrm{Res}\mathrm{Ind}V$, namely the one given by $v\mapsto 1\otimes v$, where $\mathrm{Res}\mathrm{Ind}V$ is seen as the $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$).

Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).

We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. Now, we can see the representation $\mathrm{Res}\mathrm{Ind}V$ as the left $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$. Then, there is a canonical $H$-equivariant injection $V\to \mathrm{Res}\mathrm{Ind}V$ given by $v\mapsto 1\otimes v$, and there is a canonical $H$-equivariant projection $\mathrm{Res}\mathrm{Ind}V\to V$ given by $g\otimes v\mapsto gv$ for $g\in H$ and $g\otimes v\mapsto 0$ for $g\not\in H$. This projection splits the injection, and therefore the representation $V$ is canonically a direct summand of the representation $\mathrm{Res}\mathrm{Ind}V$. Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.

"Canonically" means "canonically with respect to $U$ and $V$ and kind-of canonically with respect to $G$ and $H$" here. "Kind-of canonically with respect to $G$ and $H$" means that it's functorial with respect to maps which preserve both "lying in $H$" and "not lying in $H$", and I think we can't do better. As opposed to Robin's and Marty's proof, we don't need to rely on some randomly chosen system of representatives of cosets or double cosets.