How does the kernel of the map $\Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)$ relate to equivariant cohomology? This question may be trivial for experts. Consider a (compact, connected) smooth manifold $X$ and a (compact connected) Lie group $G$ act on $X$. Then we have the action map
$$
\mu: G\times X\rightarrow X
$$
and the projection map
$$
p: G\times X\rightarrow X.
$$
We can consider the differential forms $\Omega^{\bullet}(X)$ as well as $\Omega^{\bullet}(G\times X)$. The maps $\mu$ and $p$ give the pull-back map
$$
\mu^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)
$$
and
$$
p^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X).
$$
Now let's consider the map $\delta$ defined to be the difference of $\mu^*$ and $p^*$, i.e.
$$
\delta=\mu^*-p^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X).
$$
It is easy to easy that $\ker \delta\subset (\Omega^{\bullet}(X))^G$. Moreover, for any vector field $\Theta$ on $X$ generated by the $G$-action, we have $\ker \delta\subset \ker\iota_{\Theta}$. Hence in the case that $G$ acts on $X$ freely, we know 
$$
\ker\delta=\Omega^{\bullet}(G/X).
$$
$\textbf{My first question}$ is: if the $G$-action is not free, could we describe the equivariant cohomology $H^{\bullet}_G(X)$ in terms of $\ker\delta$?
My second question may be a little bit vague: for the action $G$ on X we can construct a simplicial manifold
$$
\ldots G\times G\times X\Rrightarrow G\times X\rightrightarrows X
$$
by actions, multiplications and projections and hence we get a sequence of   differential forms
$$
\Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)\rightarrow \Omega^{\bullet}(G\times G\times X)\ldots
$$
$\textbf{My second question}$ is: is there any reference for the study of the above sequence?
 A: Let me answer the second question.
Let $\Omega^p_G(X)=\oplus_k (\Omega^k(X)\otimes \mathrm{Sym}^{p-k}(\mathfrak{g}^*))^G$ be the weight $p$ part of the Cartan complex with the differential $(d_G \omega)(v) = \iota_a(v)\omega$ for $v\in \mathfrak{g}$ (this is the usual differential in the Cartan model without the de Rham differential).
Let $X^\bullet$ be the simplicial manifold given by the nerve of the action groupoid of $G$ on $X$ and $\Omega^p(X^\bullet)$ the Cech complex as in your question.
Then there is a quasi-isomorphism $(\Omega^p_G(X), d_G)\hookrightarrow \Omega^p(X^\bullet)$ which in degree zero is the naive inclusion. For $p=0$ this is just the statement that the functor of $G$-invariants for a compact Lie group is exact: $\Omega^0_G(X)=\mathcal{O}(X)^G$ while $\Omega^0(X^\bullet)$ is the standard complex computing invariants of $\mathcal{O}(X)$ under the coaction of $\mathcal{O}(G)$.
For $p=1$ it works as follows. The complex $\Omega^1_G(X)$ is $\Omega^1(X)^G\rightarrow (\mathfrak{g}^*\otimes\mathcal{O}(X))^G$. The trick is to realize that $\mathfrak{g}^*\cong \Omega^1(G)^G$ as $G$-representations, where $G$ acts on $\Omega^1(G)$ by left and right translations and I've taken invariants with respect to the left action. Like for $p=0$, by expanding $G$-invariants you recover $\Omega^1(X^\bullet)$. Some details (and the $p=2$ case) can be found in section 5.1.2 (p. 21) of this preprint: http://math.utexas.edu/~psafronov/papers/quasihamiltonian.pdf.
The map is incompatible with the de Rham differential unless $G$ is abelian. So I wouldn't expect the map to be a quasi-isomorphism with the de Rham differential turned on (like in the usual equivariant cohomology).
