Let $\pi: M \to E$ be a $G$-equivariant vector bundle and let us adopt the notation $$C^{\infty}(M,E)^{G} = \{\tilde{s} \in C^{\infty}(M,E) | \ \forall g \in G \ \tilde{s} \cdot g = \tilde{s}\}$$ where the condition is the usual equivariance of the section. For example, give the trivial vector bundle $P \times V$ over principal bundle $P$ the equivariant structure $$(p,v) \cdot g := (p \cdot g, \rho(g^{-1}v))$$ via some linear representation $\rho: G \to GL(V)$. The since the bundle is trivial, one may write an arbitrary smooth section $\tilde(s) = (p, s(p))$ for some smooth function $s: P \to V$. Then one may easily compute that
$$C^{\infty}(P,P \times V)^{G} \cong \{ s: P \to V \ \mathrm{smooth}| \ \forall g \in G \ R^{*}_{g} s = \rho(g^{-1})s \} $$
The trouble comes when trying to then make sense of the following isomorphism
$$\Omega^{r}(P,P \times V)^{G} \cong \{ \phi \in \Omega^{r}(P,V) | \ \forall g \in G \ R^{*}_{g}\phi = \rho(g^{-1})\phi \}$$
Namely, I'm not sure even what the equivariant structure is on $\Lambda^{r} T^{*} P \otimes (P \times V)$.
EDIT: So to check I've understood this, on the $r$-th exterior bundle you define pointwise this right action
$$(p, \omega) \cdot g = \color{blue}(p \cdot g, \omega \color{red}{(}(dR_{g^{-1}})_{p \cdot g} \ \cdot, \dots, (dR_{g^{-1}})_{p \cdot g} \ \cdot \color{red}{)} \color{blue}{)} $$ Then you may extend it to $\Lambda^{r} T^{*}_{p}P \otimes (P \times V)$ as $$(p, \omega \otimes v) \cdot g = \color{blue}(p \cdot g, \omega \color{red}{(}(dR_{g^{-1}})_{p \cdot g} \ \cdot, \dots, (dR_{g^{-1}})_{p \cdot g} \ \cdot \color{red}{)} \otimes \rho(g^{-1}) v \color{blue}{)} $$
The action of $G$ on $P$ induces a right action on $T^*P$ by cotangent lift, given $p\in P$, $g\cdot \alpha=R_{g^{-1}}^*\alpha\in T^*_{g^{-1}p}P$. This induces a natural right action on $\bigwedge^kT^*P$ by extension. We then get a right action on $\bigwedge^k T^*P\otimes(P\times V)\cong V\otimes \bigwedge^k T^*P$ by tensoring the representation of $G$ with our right action. $g(v\otimes \omega)=(\rho(g^{-1})v)\otimes g\cdot \omega$.