$G$-invariant tangent bundle We are given a Lie group $G$ and a smooth manifold $M$ equipped with a smooth group action $\Phi : G \times M \to M$, where we denote $\Phi_g := \Phi(g, \cdot)$ for $g \in G$. 
Consider the following equivalence relation on the tangent bundle $TM$: For $X_x \in T_x M$ and $Y_y \in T_y M$, define
$$
X_x \sim Y_y : \iff \exists g \in G : X_x = T_y \Phi_g Y_y,
$$
denoting by $T_y \Phi_g : T_y M \to T_ {\Phi_g(y)} M$ the usual tangent map.
Then, define $TM_G := TM/ \sim$ as some kind of reduction of the tangent bundle. We can equip this with a surjective map $\pi : TM_G \to M/G$ by setting $\pi([X_x]) = [x]$, and we find a vector space structure on every preimage $\pi^{-1}([x])$.
Also, since $T\Phi : G \times TM \to TM$ can be regarded as a group action on the manifold $TM$, if we have a free and proper $\Phi$, we also have a free and proper $T \Phi$, and as such a smooth manifold structure on both $M/G$ and $TM_G$ (I'm not 100% certain whether properness of $\Phi$ implies properness of $T \Phi$, this I still need to check in detail, but my gut tells me it will work out).
My question now is: Does this structure end up being a vector bundle over $M/G$ if we assume free and proper $\Phi$? Or do we need stronger conditions? With what I've got, I can't seem to write down meaningful local trivializations. Or is there a simpler construction for this bundle where the vector bundle structure comes more naturally?
This thing we constructed here is in some sense "larger" than the usual tangent bundle: If well defined, every fiber in $T(M/G)$ is of dimension $\dim M - \dim G$, and every fiber in $TM_G$ is of dimension $\dim M$, so this is definitely something different from $T(M/G)$.
I hope the question and the setting are clear, it might be that the whole idea is bollocks, but it seemed like a very natural structure to me. All help appreciated!
 A: The answer to your question is yes.
Note that $M\to M/G$ is a principal $G$-bundle. Let $p\in M/G$, and and let $U$ be a neighborhood of $p$, such that $M|_U$ is a trivial $G$-bundle over $U$. Let $s:U\to M|_U$ be a section, and let $\widetilde{U}$ denote the image of $s$ (this is not an open set in $M$, as it has the wrong dimension).
For the bundle you are after, we have $$TM_G|_U\cong TM|_{\widetilde{U}}.$$As $\widetilde{U}$ is diffeomorphic to $U$ which can be chosen to be a ball (and contractible in particular), the vector bundle on the right hand side is trivial. This shows that $TM_G\to M/G$ is locally trivial, proving that it is a vector bundle.
Furthermore, remembering that the vertical space at any point in a principal $G$-bundle is naturally identified with the Lie algebra $\mathfrak{g}$, we have $$TM|_{\widetilde{U}}=T\widetilde{U}\oplus\mathfrak{g}$$(here, we let $\mathfrak{g}$ denote the trivial vector bundle with fiber $\mathfrak{g}$). This shows that the vector bundle in question is $$TM_G=T(M/G)\oplus\mathrm{ad}M,$$
where $\mathrm{ad}M$ denotes the adjoint bundle.
As a sanity check, one may note that the fiber of this vector bundle is indeed $\dim M$-dimensional, as stated in the post.
Edit: Having checked the links included in your (Lukas Miristwhisky) comment and learned about the Atiyah algebroid for the first time, I now understand that my answer was not accurate enough. The vector bundle in your question is indeed the same as in the Atiyah algebroid. And it is isomorphic to the above-specified direct sum, but not equal to it. That is, there are many different isomorphisms, none of which can be distinguished from the others. In fact, as it seems to me, an isomorphism (in the category of vector bundles, at least) is equivalent to a principal connection on $M\to M/G$. This may explain why it is not described as a direct sum in the literature. However, it doesn't harm to remember this (these) isomorphism(s), in particular for convincing oneself that it is indeed a vector bundle (as asked originally in this post).
