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.