Relation between $G$ bundle over free $G$ space $X$ and vector bundles over $X/G$ (Atiyah) In Atiyah K theory there is a result : If $X$ is  $G$-free  $G$-vector bundles over $X$ corresponds bijectively to vector bundles over  $X/G$ by $E \to E/G$.
I need some clarity in the proof : 
Let $p: E \to X$ be a $G$-bundle. Then  $E$ is also $G$-free. We need to prove that the orbit space $E/G$ forms a vector space over $X/G$. The mapping $p': E/G \to X/G$ is $Ge \mapsto Gp(e)$. This map is continuous. 
The fiber of $E/G$ is  $(E/G)_{Gx}= \{ Ge \in E/G : Gp(e)=Gx \}$. If we are able to prove that this set is "same" as (I think I can't  use the term  'isomorphic') $E_{x}/G$ then there will be a vector space structure on the fiber $(E/G)_{Gx}$ induced from $E_{x}/G$. Are we supposed to use the fact that $E$ is free $G$ space ? then how ?
Author says that $p': E/G \to X/G$ is locally isomorphic to $p :E \to X$. Not sure about the following  proof : 
Let $U$ be an open subset of $X$. Then $\pi(U)$ is an open subset of $X/G$ as $\pi : X \to X/G $, the natural map is the quotient map. Suppose $p^{-1}(U) $  is isomorphic to $U \times V$, then $ p'^{-1}(\pi(U)) $ is isomorphic to $\pi(U) \times V$. That is how local triviality of $E$ gives the local triviality of $E/G$. Can any one suggest better methods ?
The converse part is clear to me.
 A: For your first question, the fiber $(E/G)_{Gx}$ is in bijection with $E_x$ (under the quotient map) and inherits the vector space structure from $E_x$. This choice is not unique, but if we choose to instead give $(E/G)_{Gx}$ the vector space structure from $gx$, then the identity on $(E/G)_{Gx}$ descends from the vector space isomorphism $g:E_x\rightarrow E_{gx}$ (which comes from the definition of a $G$-bundle over $X$), and therefore identifies the two vector space structures on $(E/G)_{Gx}$. This gets rid of any ambiguity in the definition of the vector space structure on $(E/G)_{Gx}$.
For your second question, you are right in your general approach to trivialization. But we need to be careful about the $U$ we pick (and use Atiyah's assumptions that $X$ is Hausdorff and $G$ is finite). Fix an arbitrary $x\in X$. By freeness of the group action, for each $g\in G$ not the identity, we have $x\neq gx$ and thus we can find neighborhoods $U_g\ni x$ and $V_g\ni gx$ with $U_g\cap V_g=\emptyset$ (using the Hausdorff-ness of $X$). Also let $U_e=V_e$ be a neighborhood of $x$ over which $E$ is trivializable (where $e\in G$ is the identity). Now, using the finiteness of $G$, define the open set $$W=\bigcap_{g\in G}(U_g\cap g^{-1}V_g).$$ Since $x\in U_g$ and $gx\in V_g$ for every $g\in G$, this $W$ is a neighborhood of $x$. Also $W\subset U_e$ and thus $E$ is trivializable over $W$.  Finally, notice that $$W\cap gW\subset U_g\cap V_g=\emptyset$$ for every $g\in G$ not the identity. This implies that $\pi:W\rightarrow\pi(W)$ is a homeomorphism, which enables the identification of the open set $W$ with the set it covers in the quotient. The analogous identification in the total space gives you the desired trivialization: $$p'^{-1}\big(\pi(W)\big)\cong p^{-1}(W)\cong W\times \mathbb C^n.$$ (Of course, there's nothing special about $\mathbb C$ here, I'm just writing it for concreteness. The bundle could also be real, quaternionic, or various other things.)
