Here, $K$ is complex K-theory, $R(G)$ is the complex representation ring of $G$ (which -- for now, though it shouldn't matter -- is a finite group), and $K_G$ is $G$-equivariant complex K-theory.
In fact, I don't even see how the map is supposed to run. If $E\rightarrow X$ is a vector bundle and $G \rightarrow \mbox{Aut}(V)$ is a $G$-representation, then I should be able to get a vector bundle out of the pair $(E,V)$. (If so, I can extend in the obvious way to formal differences in both slots.) If $\mbox{rk }E = \dim V =r$, then we can just convert $E$ into a principal $U(r)$-bundle $\mathcal{F}(E)$ and apply the Borel construction $\mathcal{F}(E) \times_{U(r)} V$; up to messing with which side things are acting on, this admits a fiberwise $G$-action and we're good to go. I'll denote this composite construction by $E \cdot V$ for brevity. If $E$ and $V$ don't have the same rank, then presumably we should stabilize until they do and then proceed as before, but I'm having trouble working this out. If $\mbox{rk }E=n$ and $\dim V=n+d$, then we could carry this out as $E\cdot V = (E\oplus\underline{\mathbb{R}^d})\cdot V - \underline{\mathbb{R}^d} \cdot V$ ... but now we're stuck trying to define the subtrahend, which is no better than before. Ditto for when the ranks are switched.
Then, why is this an isomorphism if $G$ acts trivially on $X$? I've seen this stated breezily in two different sources, with no indication of why it should be true. To describe a backwards map, we might take a $G$-equivariant bundle $E \rightarrow X$, choose a point $x\in X$ and then decompose $E \stackrel{\sim}{\leftarrow} \mathcal{F}_{E_x}(E) \times_{U(r)} E_x$ (where by $\mathcal{F}_{E_x}(E)$ I mean the frames modeled on the complex vector space $E_x$). But I'm having trouble convincing myself that this is indeed $G$-equivariant. I suspect I'm missing something basic here; I'm not sure I have a single equivariant bone in my whole body.