In page 40 of this article Equivariant cohomology with generalized coefficients" we find the following
Let $G$ be a lie group and $K$ is a closed subgroup of it. Let $M$ be a $K$-manifold. Consider the product manifold $G×M$. The group $K$ acts freely on the right on $G×M$ by $(g,m)k= (gk, k^{-1}m)$. Consider the fiber space $\mathcal{M} = G×_KM$ (over $G/K$) of orbits of the K-action. The group $G$ acts on the left on $\mathcal{M}$. If $\alpha \in \mathcal{A}(\mathcal{M}) \subset \mathcal{A}(G×M)$, and $g \in G$, then $ \alpha (g)$ is an element of ${(\bigwedge \mathfrak{g}^* \otimes \mathcal{A}(M) }_{hor K} $, where $\mathfrak{g}^*$ is identified with left invariant 1-forms on $G$. Thus $$ \mathcal{A}(\mathcal{M})= C^\infty (G, {{(\bigwedge \mathfrak{g}^* \otimes \mathcal{A}(M) }_{hor K})}^K.....(*)$$
How can we obtain the formula $ {(*)}$ ? I'm aware of the result which says that If $P \rightarrow M$ is a $G$-principal bundle and $E$ is a $G$-space and $P×_GE$ is an associated bundle on $M$ then $\mathcal{A}^q(M, P×_GE)$ is isomorphic to ${\mathcal{A}^q(P, E)}_{bas}$, but I don't know how to use it to get the formula ${ (*)}$ !
Your help would be greatly appreciated! Thank you.