Geometric Version of Mackey's Theorem

I am trying to prove the following theorem, which by goes by the name $\textit{Mackey's Theorem, geometric version}$ (32.1) in these notes.

$\textbf{Claim:}$ Let $G$ be a finite group, with $H,K \leq G$ and suppose that $\tau_{H}, \tau_{K}$ are $H$ (resp. $K$)-representations on representation spaces $V_{H}$, and $V_{K}$ respectively. Then the space of intertwining maps $\operatorname{Hom}_{G}(\operatorname{Ind}_{H}^{G}(V_{H}), \operatorname{Ind}_{K}^{G}(V_{K}))$ is linearly isomorphic to the space of functions, $\tilde{A}$, $f : G \rightarrow \operatorname{Hom}(V_{H}, V_{K})$ satisfying the condition:

$$f(kxh) = \tau_{K}(k)f(x)\tau_{H}(h) \ \forall k \in K, h \in H, x \in G$$

I'm struggling to understand the proof given in the linked notes, but I think I have a proof, using idempotents, for the special case that $\tau_{H}$ and $\tau_{K}$ are the trivial representations. This proof goes as follows.

Define $e_{H}: G \rightarrow \mathbb{C}$ such that $e_{H}(g) = \frac{1}{\left|H \right|} \chi_{H}(g)$ where $\chi_{H}$ is the characteristic function for $H$, and define $e_{K}$ similarly. Then let $A = \mathbb{C}[G]$ be the convolution algebra of functions $f : G \rightarrow \mathbb{C}$ where the convolution is defined to be

$$(f * f')(x) = \sum_{g \in G} f(xg^{-1})f'(g)$$

Then one can check that $e_{H}, e_{K} \in A$ and furthermore they are idempotents. Then one can check that $\operatorname{Ind}_{H}^{G}(\operatorname{triv}_{H}) \cong Ae_{H}$, and $\operatorname{Ind}_{K}^{G}(\operatorname{triv}_{K}) \cong Ae_{K}$, and that $e_{K}Ae_{H}$ is isomorphic the space of functions, $\tilde{A}$ say, $f : G \rightarrow \mathbb{C}$ such that $f(kgh) = f(g)$.

Then we have the result that for any associative unital algebra $A$ with idempotents $i,j$ that $\operatorname{Hom}_{A}(Ai, Aj)$ is linearly isomorphic to $jAi$, and so

$$\operatorname{Hom}_{G}(\operatorname{Ind}_{H}^{G}(\operatorname{triv}_{H}), \operatorname{Ind}_{K}^{G}(\operatorname{triv}_{K})) = \operatorname{Hom}_{A}(Ae_{H}, Ae_{K}) \cong e_{K}Ae_{H} \cong \tilde{A}$$

I tried to adapt this proof to prove the full version, with the convolution algebra $A$ being the maps $f : G \rightarrow \operatorname{Hom}(V_{H}, V_{K})$, and idempotents $e_{H} = \frac{1}{\left| H \right|}\dot{\tau}_{H}$ where

$$\dot{\tau}_{H}(g) = \left\{\begin{array}{lr} \tau_{H}(g), & \text{if} \ g \in H \\ 0, & \text{otherwise } \end{array}\right.$$ and $e_{K}$ defined similarly. But my problem now is that $e_{H}$ and $e_{K}$ are not elements of my convolution algebra $A$ (although it does make sense to pre-convolve elements of $f$ with $e_{K}$ and post-convolve with $e_{H}$), and that $e_{K}Ae_{H}$ is indeed $\tilde{A}$.

Does anyone happen to have some advice on whether or not this style of proof is workable, and if so how I might proceed?