Suppose $(M^n, g)$ and $(N^n,h)$ are Riemannian manifolds, and for each $p\in M$ there is a neighborhood $U\subset M$ and a diffeomorphism $\phi:(U, g|_U) \to (N,h)$ such that $g|_U = \phi^*h$. I only know this map exists, I don't have an explicit formula for it. Let $\nabla^g$ and $\nabla^h$ be the Levi-Civita connections of $M$ and $N$ respectively. For $u\in C^\infty(N)$ I want to evaluate the integral
$\int_N |\nabla^h u|^2 Vol_h$
by pulling everything back to $U\subset M$. I know that in this case, since $\phi(U) = N$ and $Vol_g = \phi^*(Vol_h)$ that, for example for $f\in C^\infty(M)$ we have
$\int_U f\, Vol_g = \int_{\phi^{-1}(N)} [(f \circ \phi^{-1})\circ \phi] \phi^*(Vol_h) = \int_{\phi^{-1}(N)} \phi^*[(f\circ \phi^{-1}) Vol_h] = \int_N (f\circ \phi^{-1}) Vol_h$
but I can't figure out how to pull this off with the connection in the first integral. Namely I need the $\nabla^h$ to turn into a $\nabla^g$ under $\phi$ somehow. I want to say something like $\nabla^g = \phi^*\nabla^h$ so that
$\int_N |\nabla^h u|^2 Vol_h = \int_{\phi(U)} |((\phi^{-1})^*\nabla^g)((u \circ\phi)\circ \phi^{-1})|^2 (\phi^{-1})^*(Vol_g) = \int_{\phi(U)} (\phi^{-1})^*[|\nabla^g(u\circ\phi)|^2 Vol_g] = \int_U |\nabla^g (u\circ\phi)|^2 Vol_g$
but the whole thing just looks so suspicious especially with the squared norm. I get that $|\nabla^h u|^2$ is just a smooth function on $N$ and it should pullback to M as just $|\nabla^h u|^2 \circ \phi$ but how can I use $\phi$ to transform $|\nabla^h u|^2$ into $|\nabla^g F|^2$ for some $F\in C^\infty(M)$?
It's possible to address this using pullback connections, but it isn't needed here: for any affine connection $\nabla$ and any $f\in C^\infty M$, we have $\nabla f=df$ by definition, and the differential obeys the pullback relation $\psi^*(df)=d(\psi^*f)$ for an arbitrary smooth map $\psi$.
From here, we have $\|\nabla u\|^2=\overline{h}(du,du)$, where $\overline {h}$ denotes the inverse of the metric $h$. By choosing an orthonormal basis $e_1,\cdots,e_n\in T_pM$, and using the fact that $\phi$ is an isometry, and thus the pushforwards $d_p\phi(e_1),\cdots,d_p\phi(e_n)$ form an orthonormal basis of $T_{\phi(p)}M$, we have $$\begin{align*} \overline g(d(\phi^*u),d(\phi^*u))|_p&=\sum_{i=1}^n[d(\phi^*u)(e_i)]^2 \\ &=\sum_{i=1}^n[\phi^*(du)(e_i)]^2 \\ &=\sum_{i=1}^n[du(d_p\phi(e_i))]^2 \\ &=\overline{h}(du,du)|_{\phi(p)} \end{align*}$$