About Poincaré-Hopf theorem: can we find and draw $v$ with only one zero on $S$? 
Disclaimer: It turns that that there is a positive answer to a more general question than my original question (see here), but this answer isn't what I am really looking for: I would like some visual constructions. Also I don't want to make a new post because an answer has already been posted.

The Poincaré-Hopf theorem (for surfaces) states that if $v$ is a smooth vector field with isolated zeros on a compact surface $S$, then the sum of the index of these zeros is equal to the Euler characteristic of $S$:
$$\sum_{x\text{ zero of }v} ind(v,x)=\chi(S).$$
Also during the proof that I know, we constructed explicitly a vector field $v$ on $S$ with isolated zeros, namely by triangulating the surface. My original question was:

In the case where $\chi(S)\neq 0$, i.e $S$ is neither a torus nor a klein bottle, can we ask that $v$ has only one zero $x_0$, which therefore satisfies $ind(v,x_0)=\chi(S)~?$

So it turns out that this is true in general (even in higher dimensions). What I really want to know, now that the existence of such vector field is proven, is this:

Can we provide some (nice) pictures of these vector fields $v$ on $S$?

This can be done when $S$ is the sphere $S^2$, by pulling back the constant vector field $\frac{\partial}{\partial x}$ on $\Bbb R^2$ via the stereographic projection and extending this pulled-back vector field to $0$ on the north pole. The picture is as follows (with a zoom on the right):

For the case where $S$ is the protective plane $\Bbb{RP}^2=S^2/_{\{\pm id\}}$, this is also possible. 
If we take the vector field on the sphere which is drawn below (each trajectory stays in horizontal hyperplanes basically) then this vector field has two zeros (the north pole $N$ and the south pole $S$), both of degree one. Moreover, this vector field is invariant under the antipodal map, and therefore it gives rise to a vector field on $\Bbb{RP}^2$ and this vector field has only one zero, namely $x_0=\{N,S\}$, with degree $1$.

For the case $S=M_g$ of the connected sum of $g\geq 2$ torus, I tried to do drawings and write $M_g$ as a polygon with edges identified, but I can't find an example. For example at least when $g=2$, I am hoping to find a comprehensive picture (on the embedded surface or on the polygon with edges identified).
I haven't thought about the other cases yet (connected sum of projective planes), but it should be of the same difficulty.
Any help would be greatly appreciated!
 A: 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.
