$\def\sgn{\operatorname{sgn}}
\def\sD{\mathcal{D}}$I would be grateful if someone could correct me in case there is any typo in the following expressions.
After solving the ODE, the flows for $V$ and for $W$ can be seen to be, respectively,
$$
\begin{aligned}
\theta_t(x,y,z)&=
\begin{cases}
(x+t,0,z),&y=0,t>-x,\\
(x+t,y,z-\arctan\frac{x+t}{y}+\arctan\frac{x}{y}),&y\neq 0.
\end{cases}\\
\psi_s(x,y,z)&=
\begin{cases}
(0,y+s,z),&x=0,s>-y,\\
(x,y+s,z-\arctan\frac{y+s}{x}-\arctan\frac{y}{x}),&x\neq 0.
\end{cases}
\end{aligned}
$$
Denote $\sD_\theta$ and $\sD_\psi$ to the flow domains of $\theta$ and $\psi$, respectively.
Composing in the two possible orders, on the one hand we have that for all pair of times $(t,s)$ and points $p=(x,y,z)$ such that $(s,p)\in\sD_\psi$ and $(t,\psi_s(p))\in\sD_\theta$,
$$
\theta_t(\psi_s(x,y,z))
=
\begin{cases}
(t,y+s,z-\arctan\frac{t}{y+s}) & x=0,s>-y,\\
(x+t,0,z-\arctan\frac{y}{x}) & x\neq 0, y=-s, t>-x\\
(x+t,y+s,\\z+\arctan\frac{y+s}{x}+\arctan\frac{x}{y+s}-\arctan\frac{x+t}{y+s}-\arctan\frac{y}{x})
& x\neq 0, y\neq -s.
\end{cases}
$$
On the other hand, for all times $(t,s)$ and $p=(x,y,z)$ with $(t,p)\in\sD_\theta$ and $(s,\theta_t(p))\in\sD_\psi$,
$$
\psi_s(\theta_t(x,y,z))
=
\begin{cases}
(x+t,s,z+\arctan\frac{s}{x+t}) & y=0,t>-x,\\
(0,y+s,z+\arctan\frac{x}{y}) & y\neq 0, x=-t, s>-y\\
(x+t,y+s,\\z-\arctan\frac{x+t}{y}-\arctan\frac{y}{x+t}+\arctan\frac{y+s}{x+t}+\arctan\frac{x}{y})
& y\neq 0, x\neq -t.
\end{cases}
$$
Recall that for $u\in\mathbb{R}\setminus\{0\}$, we have
$$
\label{1}\tag{1}
\frac{\pi}{2}\sgn u=\arctan u+\arctan\frac{1}{u}.
$$
(This identity follows after differentiating by $u$.)
The only possible choice of $(x,y,z,t,s)$ such that both $\theta_t(\psi_s(x,y,z))$ and $\psi_s(\theta_t(x,y,z))$ are defined and such that at the same time $\theta_t(\psi_s(x,y,z))\neq\psi_s(\theta_t(x,y,z))$, is for $-t\neq x\neq 0\neq y\neq -s$. Indeed, if $(x,y,z,t,s)$ are as such, then both orders of flow compositions are defined and, using \eqref{1}, we have $\theta_t(\psi_s(x,y,z))=\psi_s(\theta_t(x,y,z))$ if and only if
$$
\sgn x\sgn(y+s)+\sgn y\sgn(x+t)=\sgn x\sgn y+\sgn(x+t)\sgn(y+s).
$$
But this identity can be seen to not hold for $(x,y,t,s)=(1,-1,-2,2)$.
What is happening? Consider the characterization from Lee's book of commutativity of vector fields vs commutativity of the flows (it's given here). Then, for $(x,y)=(1,-1)$ even though both $\theta_t(\psi_s(x,y,z))$ and $\psi_s(\theta_t(x,y,z))$ are defined for $(t,s)=(-2,2)$, it happens that there are no open intervals $J,K$ containing $0$ with $-2\in J$, $2\in K$ such that $\theta_t(\psi_s(x,y,z))$ is defined for all $(t,s)\in J\times K$ or $\psi_s(\theta_t(x,y,z))$ is defined for all $(t,s)\in J\times K$. (You can check this yourself by inspecting the formulas for the composites of the flows, trying to move continuously $(t,s)$ from $(-2,2)$ to $(0,0)$.)
Best Answer
In Lee's book, take a close look at Theorem 9.22, which has the label "Canonical Form Near a Regular Point" (it is on page 220 in the edition I'm using). It is the theorem which states that what you are trying to do is possible (i.e. find coordinates so that the chosen vector field is a coordinate vector field). This is also often called the "straightening theorem" (e.g. this Wikipedia page)
If you read the proof, it tells you how to construct such coordinates.
Start by finding a curve through the point $(1,0)$ which is not tangent to the vector field near the point. For example, we can take the vertical line $x=1$. We can parameterize this line using the $y$-coordinate. For each point on this vertical line, we can apply the flow of $W$ for time $t$ (for $t \in (-\varepsilon, \varepsilon)$), and this will generate a small neighborhood around $(1,0)$.
In other words, our new coordinate system is $(t,v)$, which is the point obtained by starting at $(1,v)$ (i.e. a point on the vertical line, in the original $x,y$ coordinates) and flowing for time $t$, using the $\theta_t$ you mentioned. This gives the coordinate transformation
$$ (x,y) = (e^t, v \, e^{2t}) $$
You can confirm (using the chain rule) that $\frac{\partial}{\partial t}$ is the desired vector field:
$$ \frac{\partial}{\partial t} = \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} = e^t \, \frac{\partial}{\partial x} + 2v e^{2t} \, \frac{\partial}{\partial y} = x \, \frac{\partial}{\partial x} + 2y \, \frac{\partial}{\partial y} $$