I'm trying to show that for any one-parameter family of geodesics $\gamma(s,t)$ (where $\gamma(s_0,t)$ is a geodesic for any constant $s_0 \in (-\epsilon, \epsilon)$) defined on a Riemannian manifold $M$, $J=\cfrac{\partial \gamma(s,t)}{\partial s}$ is a Jacobi field; that is, $\cfrac{D^2}{dt^2}J(t) + R\left(J(t), \frac{\partial \gamma(0,t)}{\partial t}\right)\frac{\partial \gamma(0,t)}{\partial t}=0$, where $D$ denotes the covariant derivative associated with the metric on $M$.
The book I have been reading offers a proof as follows, \begin{equation} \cfrac{D^2}{dt^2}J(t) = \cfrac{D}{dt} \cfrac{D}{dt} \cfrac{\partial \gamma(s,t)}{\partial s} \end{equation} \begin{equation}=\cfrac{D}{dt} \cfrac{D}{ds} \cfrac{\partial \gamma(s,t)}{\partial t}, \end{equation} by the symmetry of the connection defined on $M$. \begin{equation} =\cfrac{D}{ds} \cfrac{D}{dt} \cfrac{\partial \gamma(s,t)}{\partial t} + R\left( \frac{\partial \gamma(s,t)}{\partial t},J(t)\right)\frac{\partial \gamma(s,t)}{\partial t} \end{equation} \begin{equation} =R\left( \frac{\partial \gamma(s,t)}{\partial t},J(t)\right)\frac{\partial \gamma(s,t)}{\partial t}, \end{equation} as $\gamma(s,t)$ is a geodesic.
I find the second equality to be suspect. Symmetry of the connection would imply that $\cfrac{D}{dt} \cfrac{\partial \gamma(s,t)}{\partial s}-\cfrac{D}{ds} \cfrac{\partial \gamma(s,t)}{\partial t}=\left[\cfrac{\partial \gamma(s,t)}{\partial t},\cfrac{\partial \gamma(s,t)}{\partial s}\right]$, and I see no reason why the Lie bracket term would be zero. The Lie bracket of coordinate vector fields is zero, I know, but we don't really know anything about the coordinate system of $M$. If we are using a Riemannian normal coordinate system, then $\cfrac{\partial \gamma(s,t)}{\partial t}$ is a coordinate vector field, but not $\cfrac{\partial \gamma(s,t)}{\partial s}$. Am I missing something? Or are additional conditions required to make the Lie bracket zero?
For a general one-parameter family of geodesics, the Lie bracket expression you wrote down has no meaning. The way to prove this, instead, is to choose smooth local coordinates $(x^i)$ on $M$, and write $\gamma$ locally as $\gamma(s,t) = (x^1(s,t),\dots,x^n(s,t))$. Then when you expand out $$ \cfrac{D}{dt} \cfrac{\partial \gamma(s,t)}{\partial s}-\cfrac{D}{ds} \cfrac{\partial \gamma(s,t)}{\partial t} $$ in coordinates, you get an expression involving $$ \frac{\partial^2 x^k(s,t)}{\partial t\,\partial s} - \frac{\partial^2 x^k(s,t)}{\partial s\,\partial t}, $$ which is clearly zero.
This is carried out in detail in my book Riemannian Manifolds, Lemma 6.8.