The value of $\bar{\nabla}_\bar{X}\bar{Y}$ along $M$ doesn't depend on the extension $\bar{Y}$ which we choose for $Y$. Let $M$ be a submanifold of $\bar{M}$, and suppose that $X, Y \in \Gamma^\infty(TM)$ are vector fields tangent to $M$, with local extensions to $\bar{X}, \bar{Y}$ to $\bar{M}$. Then the value of $\bar{\nabla}_\bar{X}\bar{Y}$ along $M$ does not depend on the extension $\bar{Y}$ which we choose for $Y$. 
Does anyone have any insight or a proof as to why the last sentence there is true? Note that we define $\nabla_XY = (\bar{\nabla}_\bar{X}\bar{Y})^T$. 
 A: Here's an example that I hope will help. Consider the $x$-axis as our $M$ a submanifold of $\mathbb R^2$, which will be our $\bar M$. A vector field tangent to $M$ associates to each point $p = (x,0)$ in $M$, a vector $\vec v$ in $T_p\bar{M}$ that lives inside $T_pM$, viewed as a vector subspace of $T_p\bar M$. So in this case, think about assigning to each point on the $x$-axis a vector pointing negatively or positively in the $x$-direction. I'm not sure about your favourite notation, but in mine, this means I have smooth functions $f_X,f_Y\colon \mathbb R \to \mathbb R$ and vector fields $X$, with $X_p = f_X(p)\partial_px$, $Y_p = f_Y(p)\partial_px$.
To extend $X$ and $Y$ to a neighbourhood (in $\mathbb R^2$) of some open interval on the $x$-axis involves choosing a way to extend our functions $f_X$ and $f_Y$ to be defined on a two-dimensional neighbourhood, and choosing $g_X\partial y$ and $g_Y\partial_y$ on that same neighbourhood so that when $p = (x,0)$ inside this neighbourhood, $g_X(p)$ and $g_Y(p)$ are zero.
As you can see, we have an infinite number of possible choices! However, when we just look "along $M$," i.e. when we try to calculate $\nabla_XY$, we are only caring about how the $x$-coordinate varies. The coefficient $\partial_y$ in this case will always be zero, since $X$ and $Y$ are tangent to the $x$-axis, and the value of the $\partial_x$ coefficient will only depend on the values it takes on along the $x$-axis, i.e. will only depend on which $X$ and $Y$ we start with, not on the choices we made to define $\bar X$ and $\bar Y$.

This is not a proof, but the idea is that up to locally choosing an appropriate chart and varying the dimensions of $M$ and $\bar M$, this is the picture for what happens in general.
