Lie brackets of vector fields along a geodesic to obtain the Jacobi equation I have done this question in mathstack but someone has suggested me it is more appropriate to ask this here.
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?
 A: Your reasoning is correct, but I think you might be missing two cancelling terms in the Lie bracket, so I will rewrite it again.
It is convenient to set the bracket to act on something, say a function $v$. (It is good practice not to use the same letter twice for dummy indices)
\begin{align}
[\partial_t \Gamma,\partial_s \Gamma]=&\Big[\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}, \frac{\partial x^\beta}{\partial s}\frac{\partial}{\partial x^\beta}\Big] v\\
=& \frac{\partial x^\alpha}{\partial t}\left(\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial s}\right)\frac{\partial}{\partial x^\beta}v + \frac{\partial x^\alpha}{\partial t} \frac{\partial x^\beta}{\partial s}\left(\frac{\partial^2}{\partial x^\alpha\partial x^\beta}v\right)\\
&- \frac{\partial x^\beta}{\partial s}\left(\frac{\partial}{\partial x^\beta} \frac{\partial x^\alpha}{\partial t}\right)\frac{\partial}{\partial x^\alpha}v -\frac{\partial x^\beta}{\partial s} \frac{\partial x^\alpha}{\partial t}\left(\frac{\partial^2}{\partial x^\beta\partial x^\alpha}v\right)
\end{align}
Now cancel the second and third term as they are the same (partial derivatives commute), and rename $\alpha\to\beta, \beta\to\alpha$ in the third term to group it all as
$$ = \left(\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial s} -\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial t}\right) \frac{\partial}{\partial x^\beta}v $$
Now, use the chain rule, $\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\equiv \frac{\partial}{\partial t}$
$$ = \left(\frac{\partial^2 x^\beta}{\partial t \partial s} - \frac{\partial^2 x^\beta}{\partial s \partial t}\right) \frac{\partial}{\partial x^\beta}v  = 0$$
which vanishes again due to the fact that partial derivatives commute.
P.S.
The way I quickly remind myself of the fact $[\frac{\partial}{\partial t},\frac{\partial}{\partial s} ]=0$ is by viewing the variation geodesic as a map between manifolds, as you say $$\Gamma:[-\epsilon,\epsilon]\times\mathbb{I}\rightarrow \cal{M}$$
The domain is the manifold $[-\epsilon,\epsilon]\times\mathbb{I}$ completely covered by coordinates $s, t$, so $\frac{\partial}{\partial t},\frac{\partial}{\partial s}$ need to be coordinate vector fields, i.e., they must commute.
