In Q1, it looks like you counted $f$ wrong, it should be $f=2$ and therefore $v-e+f=1$.
Perhaps you mistakenly counted the 2-dimensional round disc at the bottom of the figure as one of the faces? If so, $M$ is a surface of revolution of the graph $y = \cos(x)$ ($-\pi < x < \pi$), and that 2-dimensional disc is not part of the surface.
Added to address the question in the edit: The graph that is being rotated is the graph of the function $y = \cos(x)$ ($-\pi \le x \le \pi$), more explicitly the set
$$\{(x,y) \mid y = \cos(x), -\pi \le x \le \pi\}
$$
That's the set which you correctly drew in your diagram.
So, for example, when $x=0$ you get $y=1$ and the point $(0,1)$ is part of the graph that is being rotated; however the point $(0,-1)$ is NOT part of the graph that is being rotated.
For another example, when $x=\pi/2$ you get $y=0$ and the point $(\pi/2,0)$ is part of the graph that is being rotated. However, the point $(\pi/2,-1)$ is NOT part of the graph that is being rotated.
In fact, no point with $y=-1$ is on the graph that is being rotated, except for the points $(\pi,-1)$ and $(-\pi,-1)$. Those two points, when rotated, give you a circle in the plane $y=-1$ which is part of the rotated graph. However, the points on the plane $y=-1$ which are on the inside of that circle are NOT part of the rotated graph.
If you had connected the two points $(\pi,-1)$ and $(-\pi,-1)$ by a straight line segment in $y=-1$, then that entire line segment would have been rotated and your rotated graph would have contained the disc in question. But, you did NOT connect those two points by a straight line segment.
Here's a direct connection between the index of a vector field and the obstruction theory definition of the Euler class. Note, I'm using the definition of the obstruction class from Wikipedia, since you link to it in the question.
Let $M$ be an $(n+1)$-manifold, because the indexing is slightly easier.
Let $V$ be a vector field on $M$ with finitely many zeroes. Triangulate $M$ such that each zero is in the interior of a different $(n+1)$-simplex, oriented so that it agrees with the orientation on $M$.
Then $V|_{M^{[n]}}$ (meaning the restriction of $V$ to the $n$-skeleton of $M$) is nonvanishing, and thus defines a partial section of $T^1M$ on the $n$-skeleton. The obstruction cochain will be an $n$-cochain with values in $\pi_n(S^n)\cong \Bbb{Z}$ (note that the orientation determines the isomorphism of $\pi_n$ of the fiber, $S^n$ with $\Bbb{Z}$).
The value of the $(n+1)$-cochain on a particular $(n+1)$-simplex $\Delta$ is given by $V(\partial \Delta)\in \pi_n(p^{-1}(\Delta))\cong \pi_n(S^n)\cong \Bbb{Z}$, where $p:T^1M\to M$ is the projection.
Now it's worth going over the details of this computation here.
If $\Delta$ doesn't contain a $0$ of $V$,
then $V(\Delta)$ fills in $V(\partial \Delta)$, so $V(\partial\Delta)$ is $0$ in $\pi_n(p^{-1}(\Delta))$, and thus the final integer value is $0$.
On the other hand, if $\Delta$ does contain a $0$ of $V$, $x$, then
by definition, the index of $V$ at $x$ is the degree of a map from $S^n\to S^n$ where the first $S^n$ is the boundary sphere of a disk $D^{n+1}$ in $M$ on which $T^1M$ is trivializable, and $V$'s only zero is $x$, and the second $S^n$ is the standard sphere in $\Bbb{R}^n$. The map is the composite
$$ \partial D^{n+1} \xrightarrow{V} \partial D^{n+1}\times (\Bbb{R}^{n+1}\setminus\{0\}) \to S^n,$$
where the second map is the projection onto $\Bbb{R}^{n+1}\setminus\{ 0\}$ followed by normalization. Note that the first map depends on a choice of trivialization, which should be chosen to be compatible with the orientation.
For convenience, note that we can dispense with requiring $T^1M$ to be trivializable, since the degree is a homotopy invariant, so it suffices that $T^1M$ is trivializable up to homotopy, which is in fact the case for any choice $D^{n+1}$. In particular, it is true for $D^{n+1}=\Delta$. Additionally, the degree of the map is usually computed with either homology (or cohomology), but in our case, the degree can also be computed with homotopy groups, since for the sphere we have canonical isomorphisms $H_n(S^n)\cong \pi_n(S^n)$ by the Hurewicz theorem.
However, if we use this to translate things over, we see that the value of the obstruction cochain on $\Delta$ is the index of the vector field $V$ at $x$.
Now we're almost done.
The isomorphism of $H^{n+1}(M)$ with $\Bbb{Z}$ for a closed, connected, oriented manifold $M$ is given by evaluating an $(n+1)$-cochain at the fundamental class.
If we call the obstruction $(n+1)$-cochain $\delta$, then
the integer we want is
$$
\sum_{\Delta} \delta(\Delta) = \sum_{x} \operatorname{index}_x V.
$$
Best Answer
The given inequality for $\Bbb R P^2$ is incorrect: it should be $2 - n$ (as you compute). Indeed, $\chi(\Bbb R P^2) = 1$, as we can quickly compute using the polygonal presentation in this answer.