Yes, the Reeb field $R_w$ associated to a contact form $w$ satisfies:
1) $w(R_w)=1$ , which is basically saying that $w(R_w) \neq 0$ (then you can rescale to 1)
2) $dw(R_w, .)==0$ , so that $R_w$ "kills" every vector vield under $dw$ . Notice that a Reeb field exists by linear algebra alone: your form is defined in an odd-dimensional manifold, so that it is an antisymmetric odd-dimensional quadratic form, meaning it is degenerate, so that $dw (R_w, .)$ has a solution.
You can then set, say for the standard form $w=xdy+dz$ in $\mathbb R^3 $ , then
$dw=dydx$ , so that you want $R_w:=R_x+R_y+R_z$ and, for any generic vector field $V:=V_x+V_y+V_z$, you want :
$dydx(R_x+R_y+R_z, V_x+V_y+V_z) $ so that,
$R_xV_y-R_yV_x=0$ , with $R_z, V_z$ free variables. But I don't see how to find
a general form (nor even a single solution ) for $R_w$ from this.
Here is an attempt to describe the picture of such a vector field.
First, describe your surface as a CW-complex with precisely one $0$-cell and one $2$-cell $e_2$, with however many $1$-cells are needed to generate the fundamental group. Assuming we are not in the case $S^2$, we can assume the points on $\partial e_2$ corresponding to $0$-cell are isolated.
We can just work on the 2-cell. On each $1$-cell, choose a direction for this vector field (just like the usual pictorial notation for gluing two sides with an arrow and a name).
Pick a distinguished point $v\in\partial e_2$ that identifies with the $0$-cell, and draw chords from this $v$ to the finitely many points that also identifies with the $0$-cell. Make the vector field going out from this $v$ on these chords.
If a "triangle" (or if $\mathbb{RP}^2$ only a half-disc) has vector field flowing the same way around boundary, we extend the vector field by shrinking to this $v$ like the upper half of your middle diagram (upper half plane $\cong$ disc $\cong$ triangle).
Otherwise, the "triangle" looks like $x\geq 0, y\geq 0, x+y\leq 1$ with vector field going from $(0,1)$ to $(1,0)$ (via $(0,0)$ or not). Extend this to a vector field along curves joining $(0,1)$ to $(1,0)$.
The resulting vector field is smooth if you do it right, and only zero at the $0$-cell.
Best Answer
Hint Puncturing the sphere $\Bbb S^2$ at a point, say, $N$, leaves the space $\Bbb S^2 - \{N\} \cong \Bbb R^2$; we can make this identification explicit, e.g., with stereographic projection from the point $N$. Now, $\Bbb R^2$ admits plenty of vector fields that vanish nowhere, and some of them can be used (via that identification) to construct a vector field on $\Bbb S^2$ that vanishes only at $N$.