It is well-known that, if $M$ is a smooth manifold and $G$ is a finite group of diffeomorphisms of $M$, then for each $x\in M$ fixed by the action of $G$ there exists a coordinate system in which $G$ acts orthogonally.
Since orbifolds are locally quotients of connected open subsets of Euclidean space by finite groups of diffeomorphisms, one should expect that the previous result can be generalized in the orbifold setting, via the following statement:
Let $\mathscr X=(X,[\mathscr U])$ be an orbifold. Then, there exists an orbifold atlas $\mathscr U'$ equivalent to $\mathscr U$ in which every chart is of the form $(\widetilde U, G_x,\varphi)$, where $x\in X$, $G_x$ is the local group of $x$ and $G_x$ acts orthogonally.
However, I find this fact difficult to prove from a technical point of view. The idea is that if $(\widetilde U,G,\varphi)$ is an orbifold chart contained in $\mathscr U$, then the isotropy group $G_{\widetilde x}$ acts over $\widetilde U$ fixing $\widetilde x$, so according to our previous consideration, there exists a coordinate chart $\widetilde V\subseteq\widetilde U$ centered at $\widetilde x$ in which $G_{\widetilde x}$ acts orthogonally. Then, $(\widetilde V,G_{\widetilde x},\varphi\rvert_{\widetilde V})$ should be an orbifold chart of $X$. However
Why is $\widetilde V/G_{\widetilde x}$ homeomorphic to an open subset of $X$ via the map induced by $\varphi\rvert_\widetilde V$?
The set of all charts of the form described earlier should constitute an orbifold atlas for $X$, namely $\mathscr U'$.
Why does the local compatibility condition hold?
I can give an informal argument, as follows. Let $(\widetilde U_{\!x},G_x,\varphi_x)$, $(\widetilde V_{\!\!y},H_y,\psi_y)$ be two orbifold charts of $\mathscr U'$. Then they are constructed from charts $(\widetilde U,G,\varphi)$ and $(\widetilde V,H,\psi)$ from $\mathscr U$. If $z\in U_x\cap V_y$, then this point also belongs to $U\cap V$, so from the local compatibility of $\mathscr U$ there exists a chart $(\widetilde W,K,\xi)$ with $W\subseteq U\cap V$ and embeddings $$(\widetilde U,G,\varphi)\leftarrow(\widetilde W,K,\xi)\rightarrow(\widetilde V,H,\psi).$$ From our construction, there is an associated chart $(\widetilde W_{\!\!z},K_z,\xi_z)$ in $\mathscr U'$. But there is no reason why $W_z$ is contained inside the intersection $U_x\cap V_y$. However, we can reduce the chart choosing a sufficiently small $K_z$-stable set, so that this is the case (this follows from [S, Lemma B.1.3]). But then we can construct embeddings $$(\widetilde U_{\!x},G_x,\varphi_x)\leftarrow(\widetilde W_{\!\!z},K_z,\xi_z)\rightarrow(\widetilde V_{\!\!y},H_y,\psi_y)$$ from the previous ones, sort of completing a commutative diagram. Notice that this we have to carefully choose $\widetilde z\in\widetilde W_{\!\!z}$ such that everything works.
Could there be another (smarter/easier) way to check the charts of $\mathscr U'$ satisfy the local compatibility property?
With this, we can conclude the result, for it is trivial that the atlas $\mathscr U'$ refines the given one.
Please, I am only interested in the details of this construction. I have clear the intuitive ideas behind all of this, but I have serious doubts about the points I have highlighted above.
[S] Schmeding, A. The Diffeomorphism Group of a Non-Compact Orbifold, https://arxiv.org/abs/1301.5551v4
