Exercise to see some interpretation of the Lie bracket Let $p \in M$ and consider a coordinate chart centred on $p$. Let $\varphi_{v_1}^t$ and $\varphi_{v_2}^t$ the flows of our two vector fields. Using the chart, we define $g:\mathbb{R} \to \mathbb{R}^n$ a smooth function by
$$ g(t) = \varphi_{v_1}^t \circ \varphi_{v_2}^t (p) - \varphi_{v_2}^t  \circ\varphi_{v_1}^t (p)$$
Prove that
$$ g(0) = 0, \qquad g'(0) = 0, \qquad g''(0) = -2 [v_1,v_2](p)$$
What I have tried is the following:
Using that $\displaystyle\varphi_{x}(p,0) = p$, we have:
$$g(0) = \varphi_{v_1}^0 \circ \varphi_{v_2}^0 (p) - \varphi_{v_2}^0 \circ \varphi_{v_1}^0 (p) = \varphi_{v_1}^0 \left(\varphi_{v_2} (p,0)\right) - \varphi_{v_2}^0 \left(\varphi_{v_1} (p,0)\right) = 
  \varphi_{v_1}^0 (p) - \varphi_{v_2}^0(p) = p-p = 0$$
$$g'(t) = \dfrac{\partial}{\partial t}\varphi_{v_1}^t \left(\varphi_{v_2}^t (p)\right) \dfrac{\partial}{\partial t}\varphi_{v_2}^t (p) - 
 \dfrac{\partial}{\partial t}\varphi_{v_2}^t \left(\varphi_{v_1}^t (p)\right) \dfrac{\partial}{\partial t}\varphi_{v_1}^t (p) =$$ $$= V_1\left(\varphi_{v_1}^t \left(\varphi_{v_2}^t (p)\right)\right) V_2(\varphi_{v_2}^t (p)) - V_2\left(\varphi_{v_2}^t \left(\varphi_{v_1}^t (p)\right)\right) V_1(\varphi_{v_1}^t (p))$$.
Then: $g'(0) = V_1(p)V_2(p) - V_2(p)V_1(p) = 0$
But now, I don't know how to obtain the second derivative. And I am not sure that my above solution is correct because maybe I need to use charts, but I don't know how. If somebody could help me I will be very thankful.
 A: Be careful when differentiating this function $g$. The expression you obtained for $g'(0)$ does not make sense: $V_{1}(p)$ and $V_{2}(p)$ are tangent vectors, so you cannot multiply them. To correctly differentiate an expression like $g(t)$ that depends on $t$ in multiple places, it is useful to make a distinction as follows.
Define
$$
\alpha(t,h):=(\varphi^{t}_{v_{1}}\circ\varphi^{h}_{v_{2}})(p)
$$
and
$$
\beta(t,h):=(\varphi^{t}_{v_{2}}\circ\varphi^{h}_{v_{1}})(p),
$$
so that
$
g(t)=\alpha(t,t)-\beta(t,t). 
$
Let us now show that $g'(0)$ and $g''(0)$ are zero when considered as point derivations. For any smooth function $f$, we have
\begin{align}
g'(0)(f)&=\left.\frac{d}{dt}\right|_{0}(f\circ g(t))\\
&=\left.\frac{d}{dt}\right|_{0}(f\circ(\alpha(t,t)-\beta(t,t)))\\
&=\frac{\partial (f\circ(\alpha-\beta))}{\partial t}(0,0)+\frac{\partial (f\circ(\alpha-\beta))}{\partial h}(0,0).\label{1}\tag{1}
\end{align}
Now let's compute
\begin{align}
\frac{\partial (f\circ(\alpha-\beta))}{\partial t}(t,h)&=(df)_{\alpha(t,h)-\beta(t,h)}\left(\frac{\partial\alpha}{\partial t}(t,h)-\frac{\partial\beta}{\partial t}(t,h)\right)\\
&=(df)_{\alpha(t,h)-\beta(t,h)}\big(v_{1}(\alpha(t,h))-v_{2}(\beta(t,h)\big)\\
&=v_{1}(f)\circ\alpha-v_{2}(f)\circ\beta.\label{2}\tag{2}
\end{align}
In particular,
$$
\frac{\partial (f\circ(\alpha-\beta))}{\partial t}(0,0)=v_{1}(f)(p)-v_{2}(f)(p).
$$
Next,
\begin{align}
\frac{\partial (f\circ(\alpha-\beta))}{\partial h}(0,h)&=(df)_{\alpha(0,h)-\beta(0,h)}\left(\frac{\partial\alpha}{\partial h}(0,h)-\frac{\partial\beta}{\partial h}(0,h)\right)\\
&=(df)_{\alpha(0,h)-\beta(0,h)}\left(v_{2}(\varphi^{h}_{v_{2}}(p))-v_{1}(\varphi^{h}_{v_{1}}(p))\right)\\
&=(df)_{\alpha(0,h)-\beta(0,h)}\big(v_{2}(\alpha(0,h))-v_{1}(\beta(0,h))\big)\\
&=v_{2}(f)\big(\alpha(0,h)\big)-v_{1}(f)\big(\beta(0,h)\big).\label{3}\tag{3}
\end{align}
In particular,
$$
\frac{\partial (f\circ(\alpha-\beta))}{\partial h}(0,0)=v_{2}(f)(p)-v_{1}(f)(p).
$$
So equation \eqref{1} becomes
$$
g'(0)(f)=v_{1}(f)(p)-v_{2}(f)(p)+v_{2}(f)(p)-v_{1}(f)(p)=0.
$$
For the second derivative, we have
\begin{align}
g''(0)(f)&=\left.\frac{d^{2}}{dt^{2}}\right|_{0}(f\circ g(t))\\
&=\frac{\partial^{2} (f\circ(\alpha-\beta))}{\partial t^{2}}(0,0)+2\frac{\partial^{2} (f\circ(\alpha-\beta))}{\partial t\partial h}(0,0)+\frac{\partial^{2} (f\circ(\alpha-\beta))}{\partial h^{2}}(0,0).
\end{align}
Can you finish the computation? Use \eqref{2} and \eqref{3} throughout.
