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.

As you probably know, the complex $\Omega^\bullet(X^\bullet)$ is incompatible with the de Rham differential (i.e. it does not anticommute with the Cech differential), so I don't know how to compare it to the Cartan model.

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).