As in the problem statement, we let $\overline{\nabla}$ denote the Riemannian connection on $N$, and define $(\nabla_XY)(p) = \pi^\top[\overline{\nabla}_{X^*}Y^*(p)]$, where $\pi^\top\colon TN|_{f(U)} \to TM$ denotes the tangential projection. We aim to show that $\nabla$ is the Riemannian connection on $M$.
By the uniqueness of the Riemannian connection on $M$, it suffices to show that (1) $\nabla$ is a connection, (2) $\nabla$ is compatible with the metric, and (3) $\nabla$ is symmetric.
(1) $\nabla$ is a connection
Let $X_1, X_2, Y_1, Y_2$ be differentiable vector fields on $f(U)$ that extend to differentiable vector fields on an open subset of $N$. We note that
$$(\nabla_{X_1 + X_2}Y)(p) = (\overline{\nabla}_{X_1^* + X_2^*}Y^*)(p) = (\overline{\nabla}_{X_1^*}Y^* + \overline{\nabla}_{X_2^*}Y^*)(p)^\top = (\nabla_{X_1}Y)(p) + (\nabla_{X_2}Y)(p)$$
and
$$(\nabla_X(Y_1 + Y_2))(p) = (\overline{\nabla}_{X^*}Y_1^* + \overline{\nabla}_{X^*}Y_2^*)(p)^\top = (\nabla_XY_1)(p) + (\nabla_XY_2)(p)$$
and if $h\in C^\infty(f(U))$ is any differentiable function on $f(U)$, then
$$\begin{align*}
(\nabla_X(hY))(p) & = (\overline{\nabla}_{X^*}(hY^*))(p)^\top \\
& = \left[(\overline{\nabla}_{X^*}h)(p)Y^*(p) + h(p)(\overline{\nabla}_{X^*}Y^*)(p) \right]^\top \\
& = \left[ (Xh)(p)\,Y^*(p)\right]^\top + h(p)(\overline{\nabla}_{X^*}Y^*)(p)^\top \\
& = (Xh)(p)\,Y(p) + h(p)\,(\nabla_XY)(p),
\end{align*}$$
which shows that $\nabla$ is a connection.
(2) $\nabla$ is compatible with the metric.
This is a computation:
$$\begin{align*}
X_p\langle Y_p, Z_p \rangle_g & = X^*_p\langle Y^*_p, Z^*_p \rangle_{\overline{g}} \\
& = \overline{\nabla}_{X^*}\langle Y^*, Z^* \rangle_{\overline{g}}(p) \\
& = \langle (\overline{\nabla}_{X^*}Y^*)(p), Z^*_p\rangle_{\overline{g}} + \langle Y^*_p, (\overline{\nabla}_{X^*}Z^*)(p)\rangle_{\overline{g}} \\
& = \langle (\overline{\nabla}_{X^*}Y^*)(p)^\top + (\overline{\nabla}_{X^*}Y^*)(p)^\perp, Z_p \rangle_{\overline{g}} + \langle Y_p, (\overline{\nabla}_{X^*}Z^*)(p)^\top + (\overline{\nabla}_{X^*}Z^*)(p)^\perp \rangle_{\overline{g}} \\
& = \langle (\nabla_XY)(p), Z_p \rangle_g + \langle Y_p, (\nabla_XZ)(p)\rangle_g,
\end{align*}$$
where I've switched to subscripts $X_p = X(p)$ for a slight ease of notation. (Also, $\perp$ denotes the normal projection.)
(3) $\nabla$ is symmetric.
Finally, symmetry follows from noting that
$$\begin{align*}
(\nabla_XY)(p) - (\nabla_YX)(p) & = (\overline{\nabla}_{X^*}Y^*)(p)^\top - (\overline{\nabla}_{Y^*}X^*)(p)^\top \\
& = \pi^\top\!\left( (\overline{\nabla}_{X^*}Y^*)(p) - (\overline{\nabla}_{Y^*}X^*)(p) \right) \\
& = \pi^\top([X^*, Y^*]_p) \\
& = \pi^\top([X, Y]_p) \\
& = [X, Y]_p
\end{align*}$$
since $[X,Y]$ is tangent to $M$ whenever $X$ and $Y$ are.
A smooth manifold $M$ may be equipped with many different geometric structures. For example:
- A Riemannian manifold is a pair $(M,g)$, where $g$ is a Riemannian metric.
- A manifold-with-connection is a pair $(M, \nabla)$, where $\nabla$ is an affine connection on $M$ (not necessarily torsion-free).
It turns out that on a Riemannian manifold $(M,g)$, there is a "best" connection to choose, the Levi-Civita connection. The Levi-Civita connection is characterized by being both (1) compatible with the metric, and (2) torsion-free.
To reiterate: once a metric $g$ is chosen, there are many connections $\nabla$ we can work with, but the Levi-Civita is the best one (in a certain precise sense).
Given an affine connection $\nabla$, any whatseover, we can define its associated torsion tensor by
$$T^\nabla(X,Y) = \nabla_XY - \nabla_YX - [X,Y].$$
We say that a connection is torsion-free iff $T^\nabla = 0$. Some connections (like the Levi-Civita connection) have this property, but others don't.
Finally, given an affine connection $\nabla$, any whatsoever, and a coordinate chart, we can talk about the associated Christoffel symbols, denoted $\Gamma^k_{ij}$. Again, they depend on both the choice of connection and the coordinate chart.
It is a fact (exercise) that $\Gamma^k_{ij} = \Gamma^k_{ji}$ in every coordinate chart if and only if $\nabla$ is torsion-free.
Best Answer
Generally, suppose $M$ is a smooth $m$-dimensional submanifold of an $n$-dimensional smooth manifold $N$. Suppose $X,Y$ are vector fields on the larger manifold $N$, such that for each $p\in M$, we have $X(p),Y(p)\in T_pM$, i.e when restricted to the smaller submanifold, they are tangent to it. Now, if we take the Lie bracket in the larger manifold to get a vector field $[X,Y]:N\to TN$, then we can obviously restrict the domain to $M$, so we get $[X,Y]\bigg|_{M}:M\to TN$. The claim is that we can actually shrink the target space as well to get a mapping $M\to TM$, i.e the Lie bracket of vector fields tangent to a submanifold is again tangent to the submanifold.
A straight-forward proof is obtained by writing things out in local coordinates. Fix a chart $(U,\phi=(x^1,\dots, x^n))$ for the larger manifold $N$, which has the submanifold property for $M$, meaning $\phi(M\cap U)=\phi(U)\cap (\Bbb{R}^m\times \{0_{\Bbb{R}^{n-m}}\})$. With respect to this chart, note that the first $m$ coordinate vector fields $\left\{\frac{\partial}{\partial x^i}\right\}_{i=1}^m$, form at each point $p\in M\cap U$, a basis for the tangent space $T_pM$. Now, we can write on the domain $U$, \begin{align} X&=\sum_{i=1}^nX^i\frac{\partial}{\partial x^i}, \quad\text{and}\quad Y=\sum_{i=1}^nY^i\frac{\partial}{\partial x^i} \end{align} for some smooth functions $X^i,Y^i:U\to\Bbb{R}$, with the property that if restricted to $M\cap U$, then \begin{align} X^{m+1}=\dots =X^{n}=Y^{m+1}=\dots = Y^n=0.\tag{$*$} \end{align} This is the condition that $X,Y$ restrict to vector fields on $M$. Now, the Lie bracket of $X$ and $Y$ (restricted to $U$) is \begin{align} [X,Y]&=\sum_{i,j=1}^n\left(X^i\frac{\partial Y^j}{\partial x^i} - Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}. \end{align} If we restrict to $M\cap U$, then by $(*)$, it simplifies to \begin{align} [X,Y]\bigg|_{M\cap U}&=\sum_{i=1}^{m}\sum_{j=1}^n\left(X^i\frac{\partial Y^j}{\partial x^i} - Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}, \end{align} but now note that for $j\in\{m+1,\dots, n\}$, we have that $X^j|_{M\cap U}=Y^j|_{M\cap U}=0$, so taking the derivatives along the $\frac{\partial}{\partial x^i}$ direction for $i\in\{1,\dots, m\}$ means taking derivatives along direction tangent to $M$, so the result is zero (since the functions are zero when restricted to $M$, so differentiating along $M$ must yield $0$). So, the summation index above for $j$ can also be restricted: \begin{align} [X,Y]\bigg|_{M\cap U}&=\sum_{i=1}^m\sum_{j=1}^m\left(X^i\frac{\partial Y^j}{\partial x^i} - Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}. \end{align} This shows that the Lie bracket restricted to $M\cap U$ is spanned by the first $m$ coordinate vector fields, which are tangent to $M$, so $[X,Y]$ restricts to a vector field on $M$. Finally, since we can cover $M$ by such submanifold charts, this completes the proof.
There's also a coordinate-free proof of this fact, which uses that the Lie bracket of '$F$-related' vector fields are again $F$-related; now apply this fact with $F=\iota:M\to N$ being the inclusion map (see for example Lee's smooth manifolds text for more details).