Lie bracket in general relativity I am asking for a very simple, sketched out explanation of something I just learned today but I don't have the mathematical or physics background to make sure that at least I got the intuition or the idea correct.
It is my understanding that when using local coordinates on a manifold the Lie bracket vanishes.
At the same time, the curvature of spacetime has to do with the Lie bracket, and spacetime is curved.
Is the coordinate system in spacetime in GR a general local frame (how can this be possible (?)), or is it local (vanishing Lie bracket (?))? For the terms used here, please check this short answer.
 A: Note that the Lie bracket of the basis vectors vanishes.  In a coordinate chart $x$, the partial derivative operators $\left\{\frac{\partial}{\partial x^i}\right\}$ constitute a basis for the tangent bundle.  Each $\frac{\partial}{\partial x^i}$ is a vector field, and the Lie bracket between any two such vector fields vanishes:
$$\forall f \in C^\infty \ : \left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right]f = \frac{\partial^2 f}{\partial x^i \partial x^j} - \frac{\partial^2 f}{\partial x^j \partial x^i} = 0 \implies \left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right]=0\tag{$\star$}$$
This is due to the familiar symmetry of mixed partial derivatives from elementary analysis.  However, this certainly doesn't mean that the Lie bracket always vanishes.  For example, on $\mathbb R^2$ with coordinates $(x,y)$, we have that $\left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right]=0$ as always but e.g.
$$\left[\frac{\partial}{\partial x}, x \frac{\partial}{\partial y}\right] = \frac{\partial}{\partial y} \neq 0$$
as can be easily checked by applying the commutator to a test function as in $(\star)$.

Is the coordinate system in spacetime in GR a general local frame (how can this be possible (?)), or is it local (vanishing Lie bracket (?))?

A coordinate system induces a specific type of local frame, called a holonomic frame. In such a frame, the basis vector fields are given by partial derivative operators with respect to the corresponding coordinates, and the Lie brackets between those vector fields vanish as demonstrated above.  On the other hand, a general frame $\{\hat e_i\}$ cannot be written as partial derivative operators, and therefore the Lie brackets $[\hat e_i,\hat e_j]\neq 0$ in general.
A: 
At the same time, the curvature of spacetime has to do with the Lie bracket

It does not. If you have two vector fields $V$ and $W$, you can define new vector field $X\equiv[V,W]$ such that for every function $f$ on the manifold it satisfies $$X(f)\equiv[V,W](f)\equiv V(W(f)) - W(V(f)).$$
The curvature is present nowhere in the definition.

Is the coordinate system in spacetime in GR a general local frame?

Quoting from the answer you provided

In fact, this is a necessary and sufficient condition : if $\mathbf e_i$ is a frame whose Lie brackets vanish, then there are local coordinates $x^i$ such that $\mathbf e_i = \partial_i$.

So yes it is. Or more correctly (your statement has a category error), local coordinate system gives rise to a local coordinate vector fields which is a special case of general local frame for which Lie brackets vanish.
Again all of this has nothing to do with any curvature, you do not even need metric to be defined for these statements.
