I have the following somewhat awkward definition of covariant differentiation along a curve:
Let $S \subseteq \mathbb{R}^N$ be a smoothly and isometrically embedded manifold, and $\alpha : I \to S$ be a smooth curve. The covariant derivative along the curve $\alpha$ of a vector field $V: I \to \mathbb{R}^N$, $V(t) \in T_{\alpha(t)}S$ for each $t \in I$, is the orthogonal projection of $\dot{V}$ onto $T_{\alpha(t)}S$.
I feel like it ought to be easy to show that if I have an isometry $f: S \to \tilde{S}$, then covariant differentiation and the differential $df$ ought to commute in the obvious way, namely $ df \left( \nabla_{\dot{\alpha}} V \right) = \nabla_{df(\dot{\alpha})} \left( df(V) \right) $ but I spent the better part of the afternoon thinking about it without any success, at which point I resorted to taking charts and comparing coordinate expressions. That was easy but it felt like cheating. Is there a way to do this directly from the definition, exploiting the fact that $S$ and $\tilde{S}$ are embedded manifolds?