With reference in https://archive.org/details/GeneralRelativity/page/n82/mode/1up, where it is said $[u,v]=0$ that in this case are the vector fields below.
Given two vector fields along a geodesic $\gamma$ in a manifold given by: $$\partial_t \Gamma=\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\quad \quad \partial_s \Gamma=\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}$$
With $\gamma:\mathbb{I}\rightarrow \cal{M}$ and the geodesic variation of $\gamma$ given by $\Gamma:[-\epsilon,\epsilon]\times\mathbb{I}\rightarrow \cal{M}$ s.t. $\Gamma(0,t)=\gamma(t)$ and $\forall s\in [-\epsilon,\epsilon]$ we have that $t\rightarrow \Gamma(s,t)$ is a geodesic.
Now let $d\Gamma_p:\mathbb{R}\times\mathbb{R}\rightarrow T_{\Gamma(p)}\cal{M}$ s.t. $\partial_s\Gamma=(d\Gamma)_p(\frac{d}{ds},0)$ and $\partial_t\Gamma=(d\Gamma)_p(0,\frac{d}{dt})$.
In particular when we consider $p\in\gamma$ then $\partial_t\Gamma(0,t)=\dot \gamma(t)$ and $\partial_s\Gamma(0,t):=J(t)$.
These two last vector fields along $\gamma$ are the vector fields of my interest that I have written in coordinates considering $\Gamma(t,s)=x^{\alpha}(s,t)$. In addition I know that $\nabla_s\partial_t\Gamma=\nabla_t\partial_s\Gamma$ and obviously $\nabla_t\partial_t\Gamma=0$, since $\gamma$ is a geodesic.
$\textbf{I want to prove that $[\partial_t \Gamma,\partial_s \Gamma]=0$}$, where $\textbf{[}\quad\textbf{]}$ represent the Lie brackets.
To do this I have developed the following idea: $$[\partial_t \Gamma,\partial_s \Gamma]=\Big[\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}, \frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\Big]=\Big(\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial s}-\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial t}\Big)\frac{\partial}{\partial x^\beta}=\Big(\frac{\partial^2 x^\beta}{\partial t\partial s}-\frac{\partial^2 x^\beta}{\partial s\partial t}\Big)\frac{\partial}{\partial x^\beta}=0$$ $\textbf{My doubt: }$I am not sure of the second and above all of the third equality, where I have thought to write $\displaystyle\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial t}=\frac{\partial^2 x^\beta}{\partial s\partial t}$.
Do you think what I have done it is correct? If not can you tell me where there are the mistakes and how can I prove that the lie brackets give me a null vector in this case?