Recover Covariant Derivative from Parallel Transport

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:

Theorem 4.11. (Parallel translation) Given a curve $\gamma:I\rightarrow M,~t_0\in I$, and an [arbitrary] vector $V_0\in T_{\gamma(t_0)}M$, there exists a unique parallel vector field $W$ along $\gamma$ s.t. $W(t_0):=W(\gamma(t_0))=V_0$

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

If $\gamma:I\rightarrow M$ is a curve and $t_0,~t_1\in I$, the parallel translation defines an operator $$P_{t_0t_1}:T_{\gamma(t_0)}M\rightarrow T_{\gamma(t_1)}M$$ with $P_{t_0t_1}V_0=W(t_1)$, where $V_0\in T_{\gamma(t_0)}M$ and $W$ is the parallel vector field from the theorem.

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.