let $X,Y:M\to TM$ be vector fields on $M$. $\nabla_XY$ is the change in $Y$ along the flow curves of $X$. so for a point $p \in M$ let $\phi^X(t):\mathbb{R}\to M$ be a flow curve of $X$ passing through $p$ : $$(\phi^X)'=X \; \; ; \; \; \phi^X(0)=p$$ since I can't measure the change in $Y$ at two different points directly. I'll define a parallel transport map $\Pi_{tX}:T_pM\to T_{\phi^X(t)}M$ taking a vector at $p$ to its equivalent at $\phi^X(t)$. Then the derivative at $p$ is: $$\nabla_XY|_p=\lim_{t\to 0}\frac{\delta_XY|_{\phi^X(t)}}{t}=\lim_{t\to 0}\frac{Y_{\phi^X(t)}-\Pi_{tX}(Y_p)}{t}$$ I tried to do the usual approximation trick by assuming that for a "small" enough $t$ : $$\nabla_XY|_p \approx \frac{\delta_XY|_{\phi^X(t)}}{t}$$ but there is a problem that LHS and RHS are not at the same point.

here is what I did "correct me if i'm wrong":

since the points $p$ and $\phi^X(t)$ are so close: $$\Pi_{tX}(\nabla_XY|_p) \approx \frac{\delta_XY|_{\phi^X(t)}}{t}$$ in other words: $$\Pi_{tX}(\nabla_XY|_p)=\frac{\delta_XY|_{\phi^X(t)}}{t}+\vec\epsilon_1$$ $\vec\epsilon$ being a really small difference vector. Another thing that follows is that the values of the vector field $\nabla_XY$ at the two points are close so: $$\Pi_{tX}(\nabla_XY|_p)=\nabla_XY|_{\phi^X(t)}+\vec\epsilon_2$$ then $$\delta_XY|_{\phi^X(t)}=t\nabla_XY|_{\phi^X(t)}+t\vec\epsilon_2-t\vec\epsilon_1$$ since $\vec\epsilon$ errors are already small. $t\vec\epsilon$ would be even much smaller "second order error" and could be neglected: $$\delta_XY|_{\phi^X(t)} \approx t\nabla_XY|_{\phi^X(t)}$$ so: $$\Pi_{tX}(Y_p) \approx (-t\nabla_XY+Y)|_{\phi^X(t)}$$

so is there anything wrong with this attempt?