The lecture notes I used in my differential geometry class defined attaching to $P$ the fiber $V$ (w.r.t $P$) as follows:
We define the vector bundle obtained by attaching to $P$ the fiber $V$ (w.r.t. $\rho$ ) as the resulting quotient $$ E(P, V)=E(P, V,\rho):=(P \times V) / G=\{[p, g]: p \in P, v \in V\} $$ where $[p, v]$ denotes the element in the quotient represented by $(p, v) \in P \times G$ (so that, in $E(P, V),[p g, v]=$ $[p, g v]$ for all $p \in P, v \in V, g \in G)$. Further, we endow $E(P, V)$ with the projection $$ \pi_E: E(P, V) \rightarrow M, \pi_E(p, v)=\pi_P(v) . $$ and the fibers will be the vector spaces: $$ E(P, V)_x=\left\{[p, v]: p \in P_x, v \in V\right\}, \quad[p, v]+[p, w]=[p, v+w] $$ In this general framework, the representation $\rho$ encodes a linear action of $G$ on $V$ from the left, $$ G \times V \rightarrow V, \quad(g, v) \mapsto g \cdot v:=\rho(g)(v), $$ which, combined with the right action of $G$ on $P$, induces a (right) action of $G$ on $P \times V$ : $$ (p, v) \cdot g:=\left(p g, g^{-1} v\right) $$
Edit: When I try to apply this to $Fr(E) \times_{\rho} V^*$, where $\rho : GL_r \to GL_r(V) : A\to (A^{-1})^t$. This is quite confusing. This space is supposed to be isomorphic to $E^*$, which means that $$[e_x\cdot A, v^*] \sim [e_x, (A^{-1})^t v^*]$$ and they should give rise to the same linear functional in $E^*_x$. Note that $v^* : V \to \mathbb{R}$. With identification $E_x \simeq_f V$ (which is not canonical (?), choose the basis as $e_x$ ) Then $v^* \circ f : E_x \to \mathbb{R}$. How do I show that $$[e_x\cdot A, v^*] \sim [e_x, (A^{-1})^t v^*]$$ give rise to the same linear functional in $E^*_x$ ?