The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $\tilde M$ its universal cover such that for any $\tilde x\in \tilde M$, write $x=p(\tilde x)$ where $p$ is the covering map. Consider a chart $x\in U$ such that there exists $\tilde U$ which contains $\tilde x$ such that the restriction $p_{\tilde U}:\tilde U\rightarrow U$ is an homeomorphism $\phi_{\tilde U}=\phi_U\circ p_{\tilde U}$.

To construct the developing map, one fixes $\tilde x_0\in \hat M$ for every $\tilde x\in \hat M$, one consider a path $c:[0,1]\rightarrow \hat M$ such that $c(0)=\tilde x_0$ and $c(1)=\tilde x$, then one consider chart $(\tilde U_0,\tilde \phi_0),...(\tilde U_n,\tilde\phi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]\subset \tilde U_i$ and one sets

$D(\tilde x)=\tilde g_0...g_{n-1}\tilde\phi_n(\tilde x)$ where $g_i\in G$ such that $\tilde \phi_i\circ{\tilde\phi_{i+1}}^{-1}$ is the restriction of $g_i$ to $\tilde\phi_{i+1}(\tilde U_{i+1})$.

This shows that if we set $U_n=p(\tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}\phi_n)$  of $x$ and the chart $(\tilde U_n,g_0...g_{n-1}\tilde \phi_n)$ which answer your question.