First of all I want to apologize for the much too long post. A Lie group $G$ is acting on a smooth manifold $M$, then we define \begin{align*} T^k_G(M)= (S^\bullet \mathfrak{g}\otimes T^k_\mathrm{poly}(M))^G \end{align*}
for $\mathfrak{g}=Lie(G)$, $T^k_\mathrm{poly}(M)=\Gamma(\Lambda^{k+1}TM)$ and $S^\bullet \mathfrak{g}$ denotes the symmetric algebra generated by $\mathfrak{g}$. Note that the invariance is taken with respect to the product of adjoint action and pushforward. Additionally, the bracket $[-,-]_G$ is defined by \begin{align*} [P\otimes X,Q\otimes Y]_G=P\vee Q\otimes [X,Y] \end{align*} on generators and shrinked to invariant elements, where $[-,-]$ is the usual Schouten bracket. For a basis $\{e_i\}_{i\in I}$ of $\mathfrak{g}$ with dual $\{e^i\}_{i\in I}$ we define, again on generators, \begin{align*} \partial(P\otimes X)=e^i(P)\otimes (e_i)_M\wedge X, \end{align*} where we denote by $(e_i)_M$ the fundamental vector field and $e^i$ is extended as a derivation of the symmetric product. Note that $\partial$ is a differential, can be shrinked to invariant elements and turns $T^\bullet_G(M)$ together with $[-,-]_G$ into a differential graded Lie algebra. If we assume that $G$ is a principal action, then one can show that \begin{align*} p\colon T^\bullet_G(M)\to T^\bullet_\mathrm{poly}(M/G), \end{align*} is a dgla map and a quasi isomorphism, where $p$ is the projection to symmetric degree $0$ followed by the "push-forward" to $M/G$ (invariant vector fields are projectable). One can show this by choosing a principal connection $\omega\in \Omega^1(M)\otimes \mathfrak{g}$ and defining \begin{align*} \tilde{h}(P\otimes X)= e_i\vee P\otimes \iota_{\omega^i}(X) \end{align*} which fulfills \begin{align*} (\tilde{h}\partial+\partial \tilde{h}) =(deg_v+deg_s)\mathrm{id}, \end{align*} for the symmetric (Lie algebra) degree $deg_s$ and the vertical vector field degree $deg_v$. Moreover, one can show that for the horizontal lift corresponding to $\omega$ $Hor_\omega\colon T_\mathrm{poly}^\bullet(M/G)\to T_\mathrm{poly}^\bullet(M)^G\subset T^\bullet_G(M)$ the following holds \begin{align*} \partial h +h\partial +Hor_\omega\circ p=\mathrm{id} \end{align*} for the homotopy $h$ constructed via $\tilde{h}$ by dividing by the suitable degrees. By the homotopy transfer theorem, one can construct an $L_\infty$ morphism $F\colon T^\bullet_\mathrm{poly}(M/G)\to T_G(M)$ with taylor coefficients $F_k\colon \Lambda^k T^\bullet_\mathrm{poly}(M/G)[1]\to T_G(M)[1-k] $. Note that the $L_\infty$ structure on $T_\mathrm{poly}(M/G)$ is given by the zero differential and the Schouten-bracket. It is easy to see (by the formula given by the homotopy transfer theorem) that $F_1$ is given by the horizontal lift $Hor_\omega$ and \begin{align*} F_2(X,Y)= e_i\otimes \Omega^i_{\alpha\beta} \iota_{dx^\alpha} Hor_\omega(X)\wedge\iota_{dx^\beta}Hor_\omega(Y) \end{align*} for the curvature 2-form $\Omega\in \Omega^2(M)\otimes \mathfrak{g}$. One can show, by a painful computation, that $F_3$ is vanishing by the Bianchi identity. Now my question(s):

1)Is this construction known? If yes , where can I find it?

2)(becomes redundant, if the answer to 1) is yes) I conjecture that the $F_k$'s vanish for $k>2$, but I cannot prove it. At least the methods I used to proof that $F_3$ do not work for $F_4$ ( I am aware that, if $F_4$ vanishes all the higher terms vanish as well), so I think that there is a more elegant/conceptual way to to show $F_3$ vanishes. Does anyone have any idea how to attack the problem? I suspect that there is some Frölicher-Nijenhuis calculus going on, which makes that trivial.

Probably, a last remark. In my mind $T_G(M)$ is the "dual" to Cartan model for equivariant cohomology, but I am not sure that this helps for finding a solution.