It is well known that one can recover the connection from the parallel transport. I struggle to understand this concept.
Since $\Gamma(\gamma)^t_s:E_{\gamma(s)}\to E_{\gamma(t)}$ is an isomorphism between the fibers, it holds that $\Gamma(\gamma)^t_s(e_s)=e_t$ for the unique parallel translate $e_t=e(\gamma(t))\in E_{\gamma(t)}$ of a section $e$ in the vector bundle along a curve $\gamma:I\to M$ . Consequently, $\nabla_{\dot{\gamma}(0)}e=\lim_{h\to 0}(\Gamma(\gamma)^0_h(e_h)-e_0)/h=0$.
However, how do we "recover" $\nabla_{X}e$ for an arbitrary section $e$ that might not be parallel along $\gamma$ where $\dot{\gamma}(0)=X$?
Best Answer
Although he discusses it for the tangent bundle only, John M. Lee mentions a theorem in his book Riemannian Manifolds, from which it should get clearer:
Here, I took the liberty of renaming the vector field.
Now, what does parallel vector field $W$ along $\gamma$ mean? Simple, it's $D_tW\equiv0$. What's $D_t$, I hear you wondering. Well, if $W$ is a smooth vector field on $M$, then $D_tW(t_0)=\nabla_{\dot\gamma(t_0)}W$. So, as not to wreak any more havoc in this definition jungle, we shall simply restrict to this case (since the ODE $D_tW\equiv0$ then becomes $\nabla_{\dot\gamma}W\equiv0$, which is the way it's written in your link).
What John does next is, he defines
This is, by the theorem, an isomorphism between tangent spaces and therefore gives you for each vector at each point of the curve a uniquely determined vector - at another point in the image of the curve - which 'looks like the original vector', if you will.
The formula would then, in John's book, look like this $$D_tV(t_0)=\lim_{t\rightarrow t_0}\frac{P^{-1}_{t_0t}V(t)-V(t_0)}{t-t_0}=\lim_{t\rightarrow t_0}\frac{W_t(t_0)-V(t_0)}{t-t_0},$$ where $W_t$ is the parallel vector field determined by $D_tW_t\equiv0$ and $W_t(t)=V(t)$.
Hope that cleared things up.