(Vishik's Normal Form) Behavior of a vector field near the boundary of a manifold

I think you are almost there. So, just a hint:

Think on the coordinate system $\tilde\varphi:V_p\to U$ satisfying $d\tilde\varphi(X)=(0,1,0)$. If you replace the first coordinate by the function $h$, you get something whose first jets at $p$ is $(ax_2,1,0)$, $a=X^2h(p)$. It shouldn't be hard to proceed from here.

The geometric meaning of the situation is that $h$ is the coordinate normal to the Torus. So, if you go in the direction of $X$, you start escaping slowly, with starting escaping velocity 0, but accelerating. Note that $Xh(p)=0$ just mean that $X$ is tangent to the torus; $X^2h(p)\neq 0$ means it is trying to escape.

ERRATA v.21.10.2018 (changing the construction from $\{0\}\times \mathbb{R}^2$ to $\{x\}\times \mathbb{R}^2$)

Actually, the diffeomorphism I suggested does not send $T^2\cap V_p$ to $\{0\}{\times} \mathbb R^2$, but the greatest issue is still the first coordinate. Let us think on the other way around:

Note that $V_p$ can be thought as a a neighborhood $U'\subset \mathbb R^3$ (up to diffeomorphism), where $T^2\cap V_p$ corresponds to $U'\cap \{0\}\times\mathbb R^2$ and the first coordinate is $h$. Let us transpose the whole problem to $U'$, still denoting by $X$ and $h$ their corresponding objects in $ U'$. Decompose $X=(X_1,X_2,X_3)$ and denote $X^T=(0,X_2,X_3)$.

The function $h$ satisfies $h(T^2)=0$, $Xh(p)=0$ and $X^2h(p)\neq 0$ (let us assume $X^2h(p)> 0$, for simplicity). The last inequality guarantees that $p$ is a regular point for the function $h_1=Xh$, therefore, $C'=h_1^{-1}(0)\cap U'$ is a regular surface transversal to $X$ and $X^T$ (you can shrink $U'$ for that). In particular, $C_x=C'\cap h^{-1}(x)$ is a smooth family of curves on $U'$ for small $x$.

Shrinking $U'$ again, we assume both $X^Th_1$ and $X^2h$ positive on $C'$.

Let $\psi:U'\to U''\subset \mathbb R^3$ be a diffeomorphism where $h$ is the first coordinate (thus preserving the first coordinate), $d\psi(X^T)=(0,1,0)$ and $z\mapsto\psi(x,0,z)$ parametrizes $C_x$ (it exists since the flow of $X^T$ preserves the first coordinate and $C_x$ is transverse to $X^T$). Now let us see what we need to change in the first coordinate. Given $(x_0,0,z_0)\in C_x$, $h_2=h\circ\psi^{-1}=first~coordinate$ satisfies

$$\psi(X)h_2(x_0,y,z+z_0)=Xh(\psi^{-1}(x_0,y,z+z_0))\\=Xh(\psi^{-1}(x_0,0,z_0))+yX^Th_1(\psi^{-1}(x_0,0,z+z_0))+z\frac{d}{dz}Xh(\psi^{-1}(x_0,0,z+z_0))+o(x_0,y,z+z_0)\\=yX^Th(\psi^{-1}(x_0,0,z+z_0))+o(x_0,y,z+z_0) $$

where $\lim_{(y,z)\to(0,0)}o(y,z)/|(y,z)|=0$ and $Xh(\psi^{-1}(0,0,z_0))=\frac{d}{dz}h(\psi^{-1}(0,y,z+z_0))=0$ since $Xh$ vanishes identicaly on $C'$.

Now we get rid of $o(x_0,y,z+z_0)$.

We just integrate $o$ on the $\psi(X)$-direction and subtract the result from $h_2$. That would give you the desired first coordinate, and if you know how to do this integration, I highly recommend to not proceed and do it yourself, since the text gets messy.

To make this integration, consider the chart $\varphi:U'''\to U''$ where $\varphi(0,0,z)\in C$ and $\varphi(t,y,z)=c(t)$, $ \dot c=\psi(X)$. Take the function $O:U''\to \mathbb R$:

$$O(x,y,z)=\int_0^{\varphi^{-1}_1(x,y,z)} o(\varphi(t,\varphi^{-1}_2(x,y,z),\varphi^{-1}_3(x,y,z)))dt.$$

If nothing went wrong, $\psi(X)(h_2-O)=\psi(X)h_2-o$, as desired.

Completing the proof:

Consider the coordinate chart $\tilde \psi:U'\to V'$ that you get by replacing the first coordinate by $h_3=h_2-O$. We have:

$$\tilde\psi(X)h_3(x_0,y,z+z_0)=Xh(\tilde\psi^{-1}(x_0,y,z+z_0))-\tilde\psi(X)O(\tilde\psi^{-1}(x_0,y,z+z_0))\\=yX^Th_1(\tilde\psi^{-1}(x_0,0,z_0))+o(\tilde\psi^{-1}(x_0,y,z+z_0))-XO(\tilde\psi^{-1}(x_0,y,z+z_0))\\=yX^Th(\tilde\psi^{-1}(x_0,0,z+z_0)). $$

(I understand that you be not happy with this term multiplying $y$. You can try to replace $O$ by another function to really finish the proof. I will try something along the week.)