Killing vector field along a geodesic

Suppose $(M,g)$ is a Riemannian manifold, $I\subseteq\mathbb R$ is an interval, and $\gamma\colon I\to M$ is a geodesic. A smooth map $F\colon I\times (-\varepsilon,\varepsilon)\to M$ is called a "variation of $\gamma$." A standard calculation shows that if $F$ is a variation through geodesics (meaning that $s\mapsto F(s,t)$ is a geodesic for each $t$), then its variation field $V(s) = \partial F(s,t)/\partial t|_{t=0}$ is a Jacobi field. (See Theorem 10.2 in my Riemannian Geometry.)

Now if $X$ is a Killing field and $\theta$ is its flow, then for each $t\in(-\varepsilon,\varepsilon)$, the diffeomorphism $\theta_t$ takes geodesics to geodesics. Thus $F(s,t) = \theta_t(\gamma(s))$ is a variation through geodesics, so its variation field $V(s) = X(\gamma(s))$ is a Jacobi field.

I continue to believe that no good sense can be made of $[X, \gamma'] = 0$. However, ignoring this, I've supplied below a proof that Killing fields are Jacobi. My method of proof was just "calculate until the desired result falls out". In particular, I offer no guarantees that there isn't a better way.

Quick conventions: My curvature tensor convention is $$ R(X, Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z $$ which makes the Jacobi equation $$ D_t^2 J + R(J, \dot{\gamma}) \dot{\gamma} = 0. $$ The condition for $X$ being a Killing field is that $$ \langle \nabla_Y X, Z \rangle = - \langle Y, \nabla_Z X \rangle \tag{1} $$ for all vector fields $Y, Z$. Setting $Y = Z$, one sees that this also implies $$ \langle \nabla_Y X, Y \rangle = 0 \tag{2} $$ for any $Y$.

Proof that Killing fields are Jacobi. Let $\Gamma$ be any extension of $\gamma'$, and let $Y$ be any vector field. It suffices to show that $$ \langle \nabla_\Gamma \nabla_\Gamma X + R(X, \Gamma)\Gamma, Y \rangle = 0 $$ for points on the geodesic.

We have \begin{align} \langle \nabla_\Gamma \nabla_\Gamma X, Y \rangle &= \Gamma \langle \nabla_\Gamma X, Y \rangle - \langle \nabla_\Gamma X, \nabla_\Gamma Y \rangle \\ &= \Gamma \langle \Gamma, \nabla_Y X \rangle - \langle \Gamma, \nabla_{\nabla_\Gamma Y} X \rangle \\ &= \langle \Gamma, \nabla_\Gamma \nabla_Y X - \nabla_{\nabla_\Gamma Y} X \rangle. \tag{3} \end{align} (In the last equation we've used $\nabla_\Gamma \Gamma = 0$, which is only true along the geodesic; I don't wish to clutter the notation any further by showing these expressions as evaluated at points $\gamma(t)$, but it is implied.)

On the other hand \begin{align} \langle R(X, \Gamma)\Gamma, Y \rangle &= -\langle R(Y, \Gamma)X, \Gamma \rangle \\ &= \langle \nabla_Y \nabla_\Gamma X - \nabla_\Gamma \nabla_Y X - \nabla_{[Y, \Gamma]} X, \Gamma \rangle. \tag{4} \end{align} Adding $(3)$ and $(4)$, and using $\nabla_\Gamma Y + [Y, \Gamma] = \nabla_Y \Gamma$, we get \begin{align} \langle \nabla_\Gamma \nabla_\Gamma X + R(X, \Gamma)\Gamma, Y \rangle &= \langle \Gamma, \nabla_Y \nabla_\Gamma X - \nabla_{\nabla_Y \Gamma} X \rangle. \tag{5} \end{align} But $$ \langle \Gamma, \nabla_Y \nabla_\Gamma X \rangle = Y \langle \Gamma, \nabla_\Gamma X \rangle - \langle \nabla_Y \Gamma, \nabla_\Gamma X \rangle = -\langle \nabla_Y \Gamma, \nabla_\Gamma X \rangle, $$ the first term vanishing because of $(2)$, whereas $$ \langle \Gamma, \nabla_{\nabla_Y \Gamma} X \rangle = -\langle \nabla_\Gamma X, \nabla_Y \Gamma \rangle $$ by $(1)$.

Thus $(5)$ reduces to zero, which is what we wanted to show.