According to Lee's book on Riemannian manifolds, a connection $\nabla$ on a smooth manifold $M$ is said to satisfy the flatness criterion if whenever $X,Y,Z$ are smooth vector fields defined on an open subset of $M$, the following identity holds: $$\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z=\nabla_{[X,Y]}Z.\tag{7.3}$$ After introducing this criterion, Lee says:
Proposition 7.2. If $(M,g)$ is a flat Riemannian or pseudo-Riemannian manifold, then its Levi-Civita connection satisfies the flatness criterion.
Its proof is shown below.
Proof. We just showed that the Euclidean connection on $\mathbb{R}^n$ satisfies (7.3). By naturality, the Levi-Civita connection on every manifold that is locally isometric to a Euclidean or pseudo-Euclidean space must also satisfy the same identity.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\square$
I think the naturality here refers to the naturality of the Levi-Civita connection on a manifold:
Proposition 5.13. Suppose $(M,g)$ and $(\widetilde{M},\widetilde{g})$ are Riemannian or pseudo-Riemannian manifolds, and let $\nabla$ denote the Levi-Civita connection of $g$ and $\widetilde{\nabla}$ that of $\widetilde{g}$. If $\phi:M\to\widetilde{M}$ is an isometry, then $\phi^*\widetilde{\nabla}=\nabla$.
But I don't know how to apply this proposition in understanding the proof of Proposition 7.2. Thank you.
Here's an answer that might be useful to math babies like me. I would only treat the Riemannian case, but the same trick should apply to pseudo-Riemannian manifolds. Let $\overline{g}$ be the Euclidean metric and let $\overline{\nabla}$ be the Euclidean connection. Since $M$ is flat, $\exists F:\mathbb{R}^n\to M$ s.t. $F$ is locally an isometry and each $p\in M$ has a neighborhood $U$ that is isometric to an open subset $V$ of $\mathbb{R}^n$. So there exists a diffeomorphism $\phi:V\to U$ such that $\phi^*g=\overline{g}$. Now pick $X,Y,Z\in\mathfrak{X}(U)$ and keep in mind that the Euclidean connection satisfies the flatness criterion. Then we would agree that $$\overline{\nabla}_{(\phi^{-1}\ )_*X}[\overline{\nabla}_{(\phi^{-1}\ )_*Y}(\phi^{-1})_*Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[\overline{\nabla}_{(\phi^{-1}\ )_*X}(\phi^{-1})_*Z]=\overline{\nabla}_{[(\phi^{-1}\ )_*X,(\phi^{-1}\ )_*Y]}(\phi^{-1})_*Z.$$ By Proposition 5.13, you know $\phi^*\nabla=\overline{\nabla}$. So, in order to show that $\nabla$ satisfies the criterion, you need to strip $\phi$ off the equation. Let's take care of each side of the equation separately. \begin{align} \text{LHS}&=\overline{\nabla}_{(\phi^{-1}\ )_*X}[(\phi^*\nabla)_{(\phi^{-1}\ )_*Y}(\phi^{-1})_*Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[(\phi^*\nabla)_{(\phi^{-1}\ )_*X}(\phi^{-1})_*Z]\\ &=\overline{\nabla}_{(\phi^{-1}\ )_*X}\{(\phi^{-1})_*\nabla_{\phi_*[(\phi^{-1}\ )_*Y]}\phi_*[(\phi^{-1})_*Z]\}\\ &\quad-\overline{\nabla}_{(\phi^{-1}\ )_*Y}\{(\phi^{-1})_*\nabla_{\phi_*[(\phi^{-1}\ )_*X]}\phi_*[(\phi^{-1})_*Z]\}\\ &=\overline{\nabla}_{(\phi^{-1}\ )_*X}[(\phi^{-1})_*\nabla_Y Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[(\phi^{-1})_*\nabla_X Z]\\ &=\text{Again!}\\ &=(\phi^{-1})_*(\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z)\\ \text{RHS}&=\overline{\nabla}_{(\phi^{-1}\ )_*[X,Y]}(\phi^{-1})_*Z\\ &=(\phi^*\nabla)_{(\phi^{-1}\ )_*[X,Y]}(\phi^{-1})_*Z\\ &=(\phi^{-1})_*\nabla_{\phi_*\{(\phi^{-1}\ )_*[X,Y]\}}\phi_*[(\phi^{-1})_*Z]\\ &=(\phi^{-1})_*\nabla_{[X,Y]}Z \end{align} Take a break and then conclude that $$\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z=\nabla_{[X,Y]}Z.$$ Thank all the comments that inspired me. You are the best!