Apply flow of $V$ to a segment of a curve, Do you get covariant derivative?

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Apply flow of $V$ to a segment of a curve, look at the velocity of the resulting new curve, a perturbation of the original curve. How is the new velocity (tangent) related to the original one (right at the start of the flow)?

Precisely:

Fix a smooth curve $\gamma$ on a manifold $M$; and a vector field $V$ defined in an open neighborhood $U \subseteq M$ of $\gamma$.

Consider the flow of $\ V$, $$ \Phi:(-\epsilon,+\epsilon)\times U \longrightarrow M, $$ which exists for small $\epsilon$.

We also write for fixed value $s$, $$\Phi _s(.):U \longrightarrow M.$$ It locally shifts $U$ to another open neighborhood of $\gamma$.

Fix $p$ on the curve, fix $s$. Then $ D\Phi _s$, the differential of $\Phi _s$, maps the tangent vector $\gamma '(p)$ to a vector at $\Phi _s(p)$.

In particular, at $s=0$, the vector $D\Phi _0(p)$ is again a vector at $p$.

Now, my question is: What is this vector? Is it the covariant derivative of $V$ along $\gamma '(p)$?

I would appreciate your shared thoughts. Thanks.

The motivation: This came up when I attempted to define a geodesic, as a path that has the shortest length among all perturbations by vectors along it. (The vector is zero at the two ends.)