This is based on my rough understanding, so let me know which part if any is wrong. Suppose $M$ is a $(G,X)$-manifold, where $X$ is a homogeneous $G$-space. Pullback a $(G,X)$ structure from $X$ to its universal cover $\widetilde{X}$. The $(G,X)$-automorphisms $\mathrm{Aut}(\widetilde{X})$ form a Lie group acting transitively on $\widetilde{X}$, making it a homogeneous $\mathrm{Aut}(\widetilde{X})$-space. There is a surjective homomorphism $\phi:\mathrm{Aut}(\widetilde{X})\to G$ pushing down automorphisms whose kernel is equal to the deck transformations of $\widetilde{X}$. The map $\widetilde{X}\to X$ is equivariant with respect to $\phi$ and the group actions from $\mathrm{Aut}(\widetilde{X})$ and $G$. Because of this any $(\mathrm{Aut}(\widetilde{X}),\widetilde{X})$-manifold is automatically a $(G,X)$-manifold. Finally, each $(G,X)$ structure on $M$ lifts to a unique $(\mathrm{Aut}(\widetilde{X}),\widetilde{X})$ structure on $M$ making the correspondence bijective.
If the argument is correct, I would appreciate it if someone could point me to a source where it is stated that $(G,X)$ structures are in one to one correspondence with $(\mathrm{Aut}(\widetilde{X}),\widetilde{X})$ structures, or something similar.
Here's a partial proof in terms of developing maps. Let's take for granted that $(\mathrm{Aut}(\widetilde{X}),\widetilde{X}))$ is a homogeneous space. Let $\mathrm{dev}:\widetilde{M}\to X$ be a developing map. From the universal property, there exists a lift $\widetilde{\mathrm{dev}}:\widetilde{M}\to \widetilde{X}$ of the developing map to the universal cover. It suffices to show the existence of a developing map $\widetilde{h}:\pi_1(M)\to \mathrm{Aut}(\widetilde{X}).$ That is, for each deck transformation $\gamma\in \pi_1(M)$ we have to prove the existence of a transformation $\widetilde{h}(\gamma)\in \mathrm{Aut}(\widetilde{X})$ such that $\widetilde{h}(\gamma)\circ \widetilde{\mathrm{dev}}=\widetilde{\mathrm{dev}}\circ \gamma$.
Let $\pi:\widetilde{X}\to X$ be the projection, and let $h:\pi_1(M)\to G$ be the holonomy representation for the $(G,X)$ structure. By the universal property, there exists an automorphism $\sigma:\widetilde{X}\to \widetilde{X}$ such that $\pi \circ \sigma = h(\gamma)\circ \pi.$ Then $$ \pi \circ \sigma \circ \widetilde{\mathrm{dev}} =h(\gamma)\circ \pi \circ \widetilde{\mathrm{dev}} =h(\gamma)\circ \mathrm{dev} =\mathrm{dev}\circ \gamma =\pi \circ \widetilde{\mathrm{dev}}\circ \gamma. $$ Since $\sigma \circ \widetilde{\mathrm{dev}}$ and $\widetilde{\mathrm{dev}}\circ \gamma$ both lift the same map, they must be related by a deck transformation $\phi$ of $\widetilde{X}$. Then $$ (\phi \circ \sigma)\circ \widetilde{\mathrm{dev}} =\widetilde{\mathrm{dev}}\circ \gamma, $$ so we can define $\widetilde{h}(\gamma)=\phi \circ \sigma$.
This $(\mathrm{Aut}(\widetilde{X}),\widetilde{X})$ structure induces the appropriate $(G,X)$ structure on $M$ because $\pi \circ \widetilde{\mathrm{dev}}=\mathrm{dev}$ by definition. Any other developing map $\widetilde{M}\to \widetilde{X}$ lifting $\mathrm{dev}$ must be separated from $\mathrm{dev}$ by a deck transformation of $\widetilde{X}$. Any deck transformation is an automorphism of $\widetilde{X}$, so this other developing map induces the same $(\mathrm{Aut}(\widetilde{X}),\widetilde{X})$ structure on $M$.