Tensorial $q$-forms of type $\rho$ on $P$ are isomorphic to $\Lambda^q(M;P\times_\rho V)$ While reading the appendix of Friedrich's "Dirac Operators in Riemannian Geometry" I met the following definition.
Let $P$ be a $G$-principal bundle over $M$, and $\rho:G\to GL(V)$ a representation.
A $q$ form  $w \in \Lambda^q(P,V)$ with values in $V$, is called tensorial of type $\rho$ if 
1) $R_g^*w = \rho(g^{-1})w$   (pseudo-tensorialty, $R_g:P\to P$ is the right  action )
2) $w_p(X_1,\dots, X_q) = 0$ if at least one of the  vectors $X_i$ is vertical i.e. $d\pi(X_i) = 0$ ($\pi:P\to M$ is the projection).
Then  the author states the following very interesting proposition without proof

The vector space of tensorial $q$-forms  of type $\rho$ on $P$ with values in $V$ is isomorphic to the vector space of $\Lambda^q(M;P\times_\rho V)$ of $q$-forms on $M$ with values in the associated vector bundle  $P\times_\rho V$.
The question is how is this isomorphism defined?

It doesn't seems straightforward to me. Given $w\in\Lambda^q(M;P\times_\rho V)$, we have something that takes tangent vectors to $M$, but we want something that takes tangent vectors to $P$, So I thought of simply pulling back $w$ using $\pi$. But the form $\pi^*w $ satisfy 2) but not 1) since $R_g^*\pi^* w = (\pi\circ R_g)^* w$ and $\pi\circ R_g = \pi$ because the right action preserve the fiber. So we would have that $R_g^*\pi^* w = \pi^* w $.
 A: As it was noted in the comments, the easy direction is going from a section $\alpha$ of the bundle $\Lambda^k(T^*P)\otimes (P\times V)$ (V valued differential forms on P) that verifies the conditions i) and ii) to a section $\alpha_M$ of the bundle $\Lambda^k(T^*M)\otimes(P\times_G V)$.
Let $\pi:P \rightarrow M$ be the projection. Once a choice of connection has been made, it induces isomorphisms fibre-wise $\pi_{*p}: H_p\subset T_pP \rightarrow T_\pi(p)M$. Via these isomorphisms one can lift vectors horizontally. So, denoting $\mathscr{p}$ the canonical projection from $P\times V$ to $P\times_G V$, $\alpha_M$ can be defined by:
$\alpha_{M,x}(X_1,\dots,X_k) = \mathscr{p}(\alpha_p(\tilde X_1,...,\tilde X_k))$
where $p$ is an arbitrary element of $\pi^{-1}(\{x\})$ and $\tilde X_i$ is obtained from $X_i$ by lifting horizontally by the procedure described above. Conditions i) and ii) guarantee that this map is well defined and independent of the choice of $p$.
The opposite direction is slightly more difficult to write down, but indeed we expect, given $\alpha_M$, to get $\alpha$ by pulling back $\alpha_M$ via $\pi$. The form $\pi^*\alpha_M$ gives is a section of $\Lambda^k(T^*P)\otimes \pi^*(P\times_G V)$, but it is not immediately clear if the vector bundle $\pi^*(P\times_G V)$ can naturally be identified with a known fibre bundle over P, namely, $P\times V$. We need to show that this is the case.
Recall that:
$\pi^*(P\times_G V) = \{ (p,\mathcal{V}) \in P \times( P\times_G V), \tilde{\pi}(\mathcal{V})=\pi(p)\}$
Where $\tilde\pi : P\times_G V \rightarrow M$ is the projection of the vector bundle $P\times_G V$.
We need to show that this bundle is trivial. This can be achieved in the following manner: Let $r\in P$; define a linear map $\tilde\rho(r)$ from $V$ into $\tilde\pi^{-1}(\{\pi(r)\})$ by $v \mapsto \mathscr{p}((r,v))$.
Note that $\tilde\rho(rg)=\tilde\rho(r)\rho(g)$ since $\mathscr{p}((rg,v))=\mathscr{p}((r,\rho(g)v))$. So we can define a map from $\pi^*(P\times_G V)$ to $P\times V$ by $(p,\mathcal{V})\mapsto (p,\tilde{\rho}(p)^{-1}\mathcal{V})$. Once we've proven that the association $r\mapsto \tilde\rho(r)$ is smooth, we've shown what we wanted.
Via this global trivialisation of $\pi^*P\times_G V$ we see that we do indeed get, via pull-back, a tensorial $V$ valued differential form. I hope this helps you !
