To derive the local expression for the affine connection. Take a vector fields $X$ and $Y$, $X$ has local expression $X= \sum_j X^j\frac{\partial}{x_j}$, $Y$ has local expression $Y= \sum_j Y^j\frac{\partial}{x_j}$, then $\nabla_XY = \nabla_{\sum_j X^j\frac{\partial}{x_j}}\sum_j Y^j\frac{\partial}{x_j}$, why does this equality hold?
My confusion is that $X= \sum_j X^j\frac{\partial}{x_j}$ only locally, and $\sum_j X^j\frac{\partial}{x_j}$ is only a vector field on an open subset of the manifold, but affine connections are for vector fields defined on the whole manifold, I do not really see why the equality would hold, it does not make any sense to me.
To be more explicit. Take charts $(U,\phi)$ of $M$, then $\forall p \in U$, $X_p = \sum_j X^j\frac{\partial}{x_j}|_p$, which is obvious, then we would consider the vector field $\sum_j X^j\frac{\partial}{x_j}$ which is defined on $U$. So we examine $\nabla_{\sum_j X^j\frac{\partial}{x_j}}\sum_j Y^j\frac{\partial}{x_j}$, the confusion arises because $\nabla$ takes vector fields defined on $M$ rather than vector fields defined on $U$ as its input.
The claim here is that in that local coordinate system, the field $\nabla_X Y$ is equal to the thing on the right. That's completely obvious, in the sense that $X$ is equal to the thing in the subscript (within this patch) and $Y$ is equal to the thing on the right (within this patch), so the left and right hand sides are equal just by substitution.
You might well ask "Well, what if we have two overlapping patches? Then we'd have two DIFFERENT expressions for $\nabla_X Y$. How do we know that they're equal?" And the answer is "That depends on how your source defines that covariant derivative." It might be that they're equal because the covariant derivative has been shown to be well-defined everywhere through some coordinate-invariant means; it might be because through explicit computation, the two different coordinate expressions have been shown equal. There are probably other proofs as well, but I can't say which one applies without knowing which definition you've used.