Suppose $P$ a principal bundle of base manifold $M$ and $P \times _\rho V$ it's associated vector bundle .We say a $V$-valued bundle is $G$-equivairant if $R_g^*\omega$=$\rho(g) \omega$,here $R_g$ denote the right mutipile on bundle $P$, more explictly, $R_g(p)=pg^{-1}$.
How to establish a bijection from $\Omega^q(M;P\times_\rho V)$ to the set of horizontal G-equivairant $V$-valued $q$-form ?
The correspondence is $\Phi\colon \Omega_{\rm hor}^q(P;V)^G \to \Omega^q(M;P\times_{\rho}V)$, given by $$(\Phi\omega)_x(v_1,\ldots, v_q) = [p, \omega_x(v_1^\uparrow,\ldots, v_q^\uparrow)]\tag{$\ast$},$$where $p \in P_x$ and the $v_k^\uparrow \in T_pP$ are lifts of $v_k$, i.e., satisfying ${\rm d}\pi_p(v_k^\uparrow)=v_k$. One has to prove that the expression on the right side of $(\ast)$ is independent on the choice of $p$ (here $G$-equivariance enters) and on the choices of lifts $v_k^\uparrow$ (here horizontality enters). Then $(\ast)$ itself suggests the formula for the inverse $\Phi^{-1}$.
You can see all the details on Differential Geometry - Connections, Curvature, and Characteristic Classes, by Loring Tu (Springer, 2017).