Lie brackets of affine vector fields

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The question comes from Goldman's text Geometric Structures on Manifolds,

Starting with an affine manifold, $M$, he defines an affine vector field, $\xi$ as a vector field on $M$ such that, for every point $p \in M$ there is a neighborhood $p \in U_{p}$ such that in coordinates we have $$\xi = \sum_{i,j=1}^{n}(a^{i}_{\ j}x^{j}+b^{i})\frac{\partial}{\partial x_{i}}$$ where $b$ is in $\mathbb{R}^{n}$ and the $a^{i}_{\ j}$ are the entries of a matrix in $\textbf{gl}(\mathbb{R}^{n})$

Goldman also has the following characterization of affine vector fields on affine manifolds with a flat affine connection:

A vector field, $X$, is affine if and only if $\nabla \nabla X = 0$ or equivalently $\nabla_{\nabla_{Y}Z}X = \nabla_{Y}\nabla_{Z} X$

The problem in question is the following:

Given an affine manifold, $M$ with a flat connection $\nabla$ that a vector field, $\xi$, is affine if and only if $[X,Y] = \nabla_{X}Y$.

The issue I am having is thus, more commonly in the definition of affine manifolds you see it stated that the manifold has a flat torsion-free connection. But then we obtain $$T(X,Y) = \nabla_{X}Y- \nabla_{Y}X - [X,Y]$$ but since the connection is torsion-free we have that $$0=\nabla_{X}Y- \nabla_{Y}X - [X,Y] $$ or $$\nabla_{X}Y- \nabla_{Y}X = [X,Y]$$

And perhaps I simply haven't dug into this deeply enough but this seems dubious to me. Like, yes clearly this holds for parallel vector fields, but the implication that $\nabla_{Y}X = 0$ seems unlikely for general affine vector fields. So I'm wondering if this is just something I need to look at in coordinates directly and prove, or more likely, that I've made some mistake and in this setting affine connections are not necessarily torsion free. I'm having a bit of difficult because passing to coordinates feels like probably the wrong play and a waste of time, but I can't tell if I'm just being squeamish here. Or whether or not there's a mistake in the book, because I can't find a reference that proves the claim in this exercise either.