For finite-dimensional complex $V$, the intertwiners $V\to W$ are exactly the fixed points of $V^{\ast }\otimes W$

For an elementary tensor $f\otimes w$ and $v\in V$, $\mathrm{\Phi }(g\cdot (f\otimes w))(v)=\mathrm{\Phi }((g\cdot f)\otimes (g\cdot w))(v)$ by [F2], and this equals $(g\cdot f)(v)(g\cdot w)=f(g^{-1}\cdot v)(g\cdot w)$ by [F1].

The right-hand side of step 1.1 is $g\cdot (f(g^{-1}\cdot v)w)=g\cdot (\mathrm{\Phi }(f\otimes w)(g^{-1}\cdot v))$, which is the value at $v$ of the conjugation action on the linear map $\mathrm{\Phi }(f\otimes w)$. Since elementary tensors span the tensor product, $\mathrm{\Phi }$ is an intertwiner of representations.

An element of $V^{\ast }\otimes W$ is fixed by $G$ exactly when its image under $\mathrm{\Phi }$ is fixed by $G$, because $\mathrm{\Phi }$ is a bijective intertwiner of step 2.1; so $\mathrm{\Phi }((V^{\ast }\otimes W{)}^G)={\mathrm{Hom}}_{\mathbb{C}}(V,W{)}^G$ by [F3].

A linear map $T$ is fixed by the conjugation action of every $g$ exactly when $g\cdot T(g^{-1}\cdot v)=T(v)$ for all $g,v$, i.e. when $T(g\cdot v)=g\cdot T(v)$; by [F4] this says precisely that $T\in {\mathrm{Hom}}_G(V,W)$. Hence ${\mathrm{Hom}}_{\mathbb{C}}(V,W{)}^G={\mathrm{Hom}}_G(V,W)$, which combines with step 3.1 into the claim.