Extending a tangent vector to a parallel vector field
Say we have a vector $Z \in T_pM$ for Riemannian manifold $(M,g)$. Lets consider a possible method for extending this vector to a parallel vector field. Let us choose coordinates $(x,y)$ centered at $p$, and first parallel translate $Z$ along the $x$ axis, and then the $y$ axis.
Using the L-C connection, it follows that $\triangledown_{\partial_y}Z=0$ for all points in our coordinate chart and that $\triangledown_{\partial_x}Z=0$ when $y=0$. We now consider the question of whether $\triangledown_{\partial_x}Z=0$ for all points in the coorindate chart. By uniqueness of parallel translates it would suffice to show that $\triangledown_{\partial_y} \triangledown_{\partial_x}Z =0$.
I don't understand why this last sentence is true. I understand that parallel translates are unique but I don't see why that shows that $\triangledown_{\partial_y} \triangledown_{\partial_x}Z =0 \rightarrow \triangledown_{\partial_x}Z=0$
This post is based on a dicussion in John Lee's Introduction to Riemannian Manifolds Chapter 7 page 117
Let me give you a hint.
Can you give one vector field $W$ that satisfies the two conditions (a) $W = 0$ for $y=0$ and (b) $\nabla_{\partial_y} W = 0$ everywhere?
Then use the fact that a vector field satisfying these two conditions is unique. So any vector field satisfying the two conditions, must be equal to the vector field $W$ you found.
The last thing to realize is that in your question $W = \nabla_{\partial_x}Z$.