Let $X$ and $Y$ be differentiable vector fields on a Riemannian manifold $M$. Let $p \in M$ and let $c:I \to M$ be a integral curve of $X$ through $p$, i.e. $c(t_0)=p$ and $\frac{dc}{dt}=X(c(t))$. Show that the Riemannian connection of $M$ is $$\nabla_XY(p)=\frac{d}{dt}(P^{-1}_{c,t_0,t}(Y(c(t))))\bigg|_{t=t_0},$$ where $P_{c,t_0,t}: T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$.
So I know that in coordinates $$\nabla_XY=\sum_{k}\left(\sum_{i,j}x_iy_j\Gamma^k_{ij}+X(y_k) \right)X_k$$ but how am I supposed to show that this horrendous equation equals $\frac{d}{dt}(P^{-1}_{c,t_0,t}(Y(c(t))))\bigg|_{t=t_0}$? I've been trying to think about this for a quite a long time now and I just don't see how we are supposed to prove this kinds of things. I know that it's probably enough to try and show this in coordinates, but these are still very unpleasant to work with.
I've faced the same problem in many instances when we need to show something related to some connection that we only have these sums and supposed to draw something out of them. Usually the solutions aren't very informative and rely on some clever choice of frames etc.
In this case I believe a clever choice of frames is necessary. The left hand side is the covariant derivative of $Y$ along $c$. Pick a chart around $p$ and a basis for $T_pM$. Translate this basis along $c$ to get a parallel frame $E_i(t)$ (defined along $c$) - this forms a basis at all points on $c$ because parallel translation is an isometry (in particular an isomorphism). Write $Y = Y^jE_j$ (along $c$) so that the parallel translate becomes $$P_{c, t_0, t}^{-1}Y = Y^j(t)E_j(t_0)\in T_pM$$ (by linearity of parallel translation and choice of the frame). So, the time derivative is $\dot{Y}^j(t_0)E_j(t_0)$ at $t_0$.
The covariant derivative of $Y^jE_j$ at $t_0$ is also the same because the covariant derivatives of $E_j$ vanish. Note that smoothness of $Y^j$s as functions of $t$ follows because everything is happening within a chart.