Proving that $\Phi_{t*}[\mathbb{Y},\mathbb{Z}]=[\Phi_{t*}\mathbb{Y},\Phi_{t*}\mathbb{Z}]$ 
Let $\mathbb{X},\mathbb{Y}$ be vector fields on $\mathbb{R}^n$. Let $\Phi_t$ denote the flow of $\mathbb{X}$. Define $L_{\mathbb{X}}\mathbb{Y}=[\mathbb{X},\mathbb{Y}]$. You are given that
$\displaystyle L^j_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}]=\sum_{k=0}^{j} \binom jk [L^k_\mathbb{X}\mathbb{Y},L^{j-k}_\mathbb{X}\mathbb{Z}]$ 
and that $\displaystyle \Phi_{t*}\mathbb{Y} = \sum_{j=0}^{\infty}\frac{(-t)^j}{j!}L^j_{\mathbb{X}}\mathbb{Y}$
Show that $\displaystyle \Phi_{t*}[\mathbb{Y},\mathbb{Z}]=[\Phi_{t*}\mathbb{Y},\Phi_{t*}\mathbb{Z}]$
Let $F$ be a diffeomorphism on $\mathbb{R}^n$, define
  $$ \mathbb{X}^i(x)=\frac{\partial F^i}{\partial x^1}(F^{-1}(x)), \mathbb{Y}^i(x)=\frac{\partial F^i}{\partial x^2}(F^{-1}(x))$$
  Show that $[\mathbb{X},\mathbb{Y}]=0$.

$\begin{align}
\displaystyle \Phi_{t*}[\mathbb{Y},\mathbb{Z}] &= \sum_{j=0}^{\infty}\frac{(-t)^j}{j!}L^j_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] \\
&= \sum_{j=0}^{\infty}\frac{(-t)^j}{j!} \left(\sum_{k=0}^{j} \binom jk [L^k_\mathbb{X}\mathbb{Y},L^{j-k}_\mathbb{X}\mathbb{Z}]\right) \\
&= \sum_{l,k=0}^{\infty} \frac{(-t)^{k+l}}{(l+k)!} \binom{l+k}{k}[L^k_\mathbb{X}\mathbb{Y},L^{l}_\mathbb{X}\mathbb{Z}] \\
&= \sum_{l,k=0}^{\infty} \frac{(-t)^{k}(-t)^{l}}{l!k!} [L^k_\mathbb{X}\mathbb{Y},L^{l}_\mathbb{X}\mathbb{Z}] \\
&= [\Phi_{t*}\mathbb{Y},\Phi_{t*}\mathbb{Z}]
\end{align}$
I cannot see how to proceed now. For the last part I assume somehow the first part has been used as I just cannot see how. 
Please that $[\cdot,\cdot]$ denotes the Jacobi bracket, not the Poisson bracket. So
$$ [\mathbb{X},\mathbb{Y}]=\left( \mathbb{X}\cdot \nabla \right) \mathbb{Y} - \left( \mathbb{Y}\cdot \nabla \right) \mathbb{X} $$ 
 A: By definition, $X=F_* \frac{\partial}{\partial x^1}$ and $Y=F_* \frac{\partial}{\partial x^2}$. If $F$ is a diffeomorphism, then $y^1:=F^1,\ldots, y^n:=F^n$ is a coordinate frame on $\mathbb{R}^n$ and $\frac{\partial}{\partial y^j}$ are commuting vector fields. But $\frac{\partial}{\partial y^i}=\sum_j \frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial x^j}=F_*(\frac{\partial }{\partial x^i})$, so $X=\frac{\partial}{\partial y^1}$ and $Y=\frac{\partial}{\partial y^2}$ commute.
I don't see how to use the previous result, as it is not obvious how to express $F_*$ as $\phi_{t*}$ for some flow.

Edit: If you want to avoid the "manifold"-notation and stay in $\mathbb{R}^n$, then define $e_1=(1,0,\ldots, 0)$ and $e_2=(0,1,\ldots,0)$ to be the Cartesian coordinate vector fields. You can easily verify that $X=F_* e_1$ and $Y=F_* e_2$. If $F=(F^1,\ldots, F^n)$ is a diffeomorphism, you can consider the components $F^1,\ldots, F^n$ to be new coordinates on $\mathbb{R}^n$ (think of polar coordinates $(r,\phi)$ on a suitable subset of $\mathbb{R}^2$ for an analogy). Then $X$ and $Y$ are the vector fields generated by first and second of these new coordinates. That means, $X$ is the field generated by keeping $F^2,\ldots, F^n$ fixed and increasing $F^1$ with unit velocity. It is a standard fact that is written in most textbook on differential geometry that such coordinate vector fields always commute.
But now I see that this is probably the core that should be proved. If you accept the notation $Xf:=df(X)$ referring to a function that in each $x$ is the derivative of $f$ in $x$ in the direction of $X(x)$, then $[X,Y]f=X(Y(f))-Y(X(f))$ and the rest follows from the simple and nice computation given by @kobe. The identity 
$Xf=(\frac{\partial}{\partial x^1}(f\circ F))\circ F^{-1}$ that he uses follows from
\begin{align}
(Xf)(x)&=(df(x))(X(x))=((\nabla f)(x))((F_* e_1)(F^{-1}(x)))=
\sum_i {\frac{\partial f(x)}{\partial x^i}} \frac{\partial F^i}{\partial x^1}(F^{-1}(x))\\
&=(\ldots \text{use the chain rule})
=(\frac{\partial}{\partial x^1} (f\circ F))(F^{-1}(x))
\end{align}
and similarly for $Y$. 
Final remark: You have probably no problem accepting $\frac{\partial^2}{\partial x^1 \partial x^2}=\frac{\partial^2}{\partial x^2 \partial x^1}$. But similarly, $\frac{\partial^2}{\partial r\,\partial\phi}=\frac{\partial^2}{\partial \phi\,\partial r}$ in polar coordinates (or any other coordinates). That's all what this is about.
A: To me, the problem does not indicate to use the first part in order to prove the second part. For the second part, note that since $X = F_{*}\frac{\partial}{\partial x^1}$ and $Y = F_{*}\frac{\partial}{\partial x^2}$, for every smooth real-valued function $f$ on $\Bbb R^n$,
\begin{align}[X,Y]f &= X(Yf) - Y(Xf)\\
&= \left(\frac{\partial}{\partial x^1}(Yf \circ F)\right) \circ F^{-1} - \left(\frac{\partial}{\partial x^2}(Xf \circ F)\right) \circ F^{-1}\\
&=\left(\frac{\partial}{\partial x^1} \frac{\partial}{\partial x^2}(f \circ F)\right) \circ F^{-1} - \left(\frac{\partial}{\partial x^2}\frac{\partial}{\partial x^1}(f\circ F)\right)\circ F^{-1}\\
&= 0
\end{align}
Therefore, $[X,Y] = 0$.
