I'm hoping someone can check that my proof of the Symmetry Lemma below is okay. (I'm actually proving a version of the result restricted to geodesic variations, because that's what I'm primarily interested in.) Note: I have seen the proof done with local coordinates, e.g., in John M. Lee's book; I want to avoid using coordinates.
Let $\gamma : [0,T] \to M$ be a geodesic in a Riemannian manifold $(M,g)$, and let $\sigma : (-\varepsilon,\varepsilon) \times [0,T] \to M$, $(s,t) \mapsto \sigma(s,t)$ be a smooth geodesic variation of $\gamma$ (i.e., $\sigma(0,\cdot) = \gamma$, and $\sigma(s,\cdot)$ is a geodesic for every $s$). We can define vector fields along $\sigma$ as follows: $$ J(s,t) = T_{(s,t)}\sigma\cdot\left.\frac{\partial}{\partial s}\right|_{(s,t)} =: \frac{\partial\sigma}{\partial s}(s,t) \quad\text{and}\quad Z(s,t) = T_{(s,t)}\sigma\cdot\left.\frac{\partial}{\partial t}\right|_{(s,t)} =: \frac{\partial\sigma}{\partial t}(s,t). $$ (Here $s$ is the coordinate on $(-\varepsilon,\varepsilon)$, and $t$ is that on $[0,T]$.) So we have $\dot{\gamma} = Z(0,\cdot)$, and $J(0,\cdot)$ is a Jacobi field along $\gamma$.
Claim: $\frac{D}{ds}Z = \frac{D}{dt}J$.
Proof: Let $\widetilde{J},\widetilde{Z} \in \mathfrak{X}(M)$ be (smooth) extensions of $J$ and $Z$, resp., to an open neighbourhood of $\sigma$, i.e., $\widetilde{J}(\sigma(s,t)) = J(s,t)$ and $\widetilde{Z}(\sigma(s,t)) = Z(s,t)$. Since $\partial/\partial s$ and $\widetilde{J}$ are $\sigma$-related, likewise $\partial/\partial t$ and $\widetilde{Z}$, we then get that $$ [\widetilde{J},\widetilde{Z}]\circ\sigma = T\sigma\cdot\left[\frac{\partial}{\partial s},\frac{\partial}{\partial t}\right] = 0. $$ Thus, since the Levi-Civita connection $\nabla$ is symmetric: \begin{align*} 0 & = T(J(s,t),Z(s,t))\\ & = T(\widetilde{J},\widetilde{Z})(\sigma(s,t))\\ & = \nabla_{\widetilde{J}}\widetilde{Z}(\sigma(s,t)) - \nabla_{\widetilde{Z}}\widetilde{J}(\sigma(s,t)) - [\widetilde{J},\widetilde{Z}](\sigma(s,t))\\ & = \nabla_{J(s,t)}(\widetilde{Z}\circ\sigma) - \nabla_{Z(s,t)}(\widetilde{J}\circ\sigma)\\ & = \frac{D}{ds}Z(s,t) - \frac{D}{dt}J(s,t). \end{align*} (End of proof.)
My main concern with this argument is the existence of the extensions $\widetilde{J}$ and $\widetilde{Z}$ (and hence that $[\widetilde{J},\widetilde{Z}]\circ\sigma = 0$). Clearly they won't exist on the entire image of $\sigma$, but if we restrict to a sufficiently small open subset $A \subset (-\varepsilon,\varepsilon) \times [0,T]$, then can we be assured that they exist on $\sigma(A)$? (Everything else in the proof seems okay to me, but I could be mistaken.)
Edit: my claim was originally written incorrectly ($J$ and $Z$ were on the wrong sides); I've corrected this.