Clarification on definition of representation of $\mathrm{Hom}(V,W)$ From Representation Theory by W. Fulton and J. Harris:
Let $V$ be a finite dimensional vector space, and $G$ a finite group.
Let $\rho: G \to \mathrm{GL}(V)$ be a representation of $V$. The dual of this representation on the dual space $V^*$ is $G\to \mathrm{GL}(V^*)$ given by $g \mapsto \Big( f \mapsto f\circ \rho(g^{-1})\Big)$.


Having defined the dual of a representation and the tensor product of two representations, it is likewise the case that if $V$ and $W$ are representations, then $\mathrm{Hom}(V, W)$ is also a representation, via the identification $\mathrm{Hom}(V, W) = V^* \otimes W$. Unraveling this, if we view an element of $\mathrm{Hom}(V, W)$ as a linear map $\varphi$ from $V$ to $W$, we have $(g \varphi)(v)=g\varphi(g^{-1}v)$.

The isomorphism $V^* \otimes W \to \mathrm{Hom}(V, W)$ is given by $f \otimes w \mapsto \Big(v \mapsto f(v)w\Big)$.
Taking into account the $G$ action we have, $g\cdot (f\otimes w) = g\cdot f \otimes g\cdot w = f\circ \rho(g^{-1}) \otimes \rho(g)(w)$ and
$$f\circ \rho(g^{-1}) \otimes \rho(g)w \mapsto \Big(v \mapsto f\circ \rho(g^{-1})(v)\cdot \rho(g)(w)\Big).$$
So, how do we have $(g \varphi)(v)=g\varphi(g^{-1}v)$?
 A: Is sufficient to check the formula $(g\varphi)(v) = g \varphi(g^{-1}v)$ when $\varphi$ is a map of the form $f(\_)w$ [since every element of $V^* \otimes W$ is a sum of elements of the form $f \otimes w$, the same happens in $\operatorname{Hom}(V,W)$].
So, suppose $\varphi = f(\_)w$, i.e., $\varphi$ corresponds with $f \otimes w$. Then observe that $g\varphi$ corresponds with $g(f \otimes w) = gf \otimes gw$, that is, $g\varphi = (gf)(\_)(gw)$. Thus
$$(g\varphi)(v) = (gf)(v)(gw) = f(g^{-1}v)(gw) \stackrel{(!)}= g f(g^{-1}v)w = g \varphi(g^{-1}v).$$
Note that, in $(!)$, we use that $a(gw) = g(aw)$ when $a$ is a scalar; in other words, left-multiplication by $g$ is linear (as the definition of representation says).
A: Let us denote the given isomorphism from $V^* ⊗ W$ to $\operatorname{Hom}(V, W)$ by $Ψ$.
You have already computed that the induced action of $G$ on $V^* ⊗ W$ is given by
$$
  g ⋅ (f ⊗ w) = ( f ∘  ρ(g^{-1}) ) ⊗ ρ(g)(w) \,,
$$
and that therefore,
$$
  ψ(g ⋅ (f ⊗ w))(v)
  =
  (f ∘ ρ(g^{-1}))(v) ⋅ ρ(g)(w) .
$$
In other words,
$$
  \tag{$\ast$}
  Ψ(g ⋅ (f ⊗ w))(v)
  =
  f(g^{-1} v) ⋅ (g w) .
$$
We note that in this formula, the multiplication sign on the right-hand side refers the scalar multiplication of the vector space $W$.
The right-hand side of $(\ast)$ tells us how $(g φ)(v)$ is supposed to look like in the special case of $φ = Ψ(f ⊗ w)$.
To get from this a formula that works for an arbitrary element of $\operatorname{Hom}(V, W)$, we need the express the right-hand side of $(\ast\ast)$ purely in terms of $φ$, $g$ and $v$, and without using $f$ or $w$.
The map $φ = Ψ(f ⊗ w)$ is given by the formula
$$
  \tag{$\ast\ast$}
  φ(v) = f(v) ⋅ w \,.
$$
By comparing the right-hand side of $(\ast\ast)$ with the right-hand side of $(\ast)$, we see that this right-hand side of $(\ast)$ can be sligthly rewritten as
$$
  g (f(g^{-1} v) ⋅ w)
$$
because the term $f(g^{-1} v)$ is a scalar, and therefore commutes with the action of $g$ on $W$.
This term can now be rewritten as
$$
  g φ(g^{-1} v)
$$
depending only on $φ$ and $g$.
