Difference between parallel transport and derivative of the exponential map

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by J. GE below.

To understand the realationship between $Q$ and $P$, it suffices to study how they act on an orthornormal(o.n.) basis. Here is the detailed argument:

Let $J_i(t)$ be the Jacobi field along $c$, with $J_i(0)=0$ and $J_i'(0)=e_i$, where $e_1, \cdots, e_{n-1}, X$ is an o.n basis of $T_p(M)$. Then one can show that $J_i(t)=Q_t(e_i)$, since $$ Q_t(e_i)=dexp_{tX} (te_i). $$

Now extend $e_i$ to $E_i(t)$ along $c$ via parallel transport $P$. One can write the $J_i$ (i.e. $Q_t$) in terms of the o.n. basis $E_i(t)$: $$ Q_t(e_i)=J_i(t)=\sum_j a_{ij}E_j(t)=\sum_j a_{ij}P_t(e_j). $$

Let $A=(a_{ij})$. Using Jacobi equation, one can show that $$ A''(t)+R(t)A(t)=0, $$ where $R(t)=(R_{ij}(t))$ is the curvature term defined by $R_{ij}(t)=\langle R(\dot{c}, E_i) \dot{c}, E_i \rangle$.

This is essentially Jacobi equation. So roughly speaking, $Q$ and $P$ satisfy the Jacobi equation.

remark: It seems my definition of $Q$ differs by a rescaling fact $t$. So if we use $N$ to denote your map, then $Q_t(e_i)= N(t e_i)$.