Here is an outline of a proof that a compact manifold $M$ with vanishing Euler characteristic has a nonvanishing vector field. I'll post it as an "answer" to provide a convenient place for comments where errors might be pointed out (or, you know, a reference given that "that's the proof that ABC used in their paper LMNOP").

Call a vector field $\vec v$ with only regular zeros (i.e. at every $p$ with $\vec v|_p = 0$, the linear map $\partial \vec v|_p : \mathrm T_p M \to \mathrm T_p M$ is invertible) *divergence-free* if for every zero $p$, $\partial \vec v|_p$ has only real eigenvalues. A zero $p$ of a divergence-free vector field $\vec v$ has a *Morse index* $\mu(p) = \#\lbrace$negative eigenvalues of $\partial\vec v|_p\rbrace$. Of course, $\operatorname{sign}(\det(\partial \vec v|_p)) = (-1)^{\mu(p)}$. By choosing a Morse function and a metric, our manifold $M$ certainly has a divergence-free vector field.

Pick some vector field $\vec v$ on $M$, and some little neighborhood without any zeros. I claim I can modify $\vec v$ by some vector field with compact support in that neighborhood to introduce two new zeros, with Morse indexes $\mu$ and $\mu+1$, for any $\mu = 0,\dots,\dim M - 1$. In one dimension, which is trivial: in a neighborhood, $\vec v = \nabla(x^3+x)$ for some coordinate $x$, and I can perturb this to $x^3 - x$. In higher dimensions, it is not much harder, and I could probably write out formulas is necessary.

Conversely, and here's the crux of the argument, where I'm not sure it's correct: Suppose I have a divergence-free vector field $\vec v$, with nearby zeros at consecutive Morse index. Then I think I can cancel them, by undoing the insertion step in the previous paragraph.

If so, then choose any divergence-free vector field $\vec v$ on $M$. Since $M$ by assumption has vanishing Euler characteristic, $\vec v$ has the same number of zeros with odd Morse index as with even Morse index. If $\vec v$ has no zeros, we're done; otherwise, choose two with Euler characteristics $\mu$ and $\mu+k$, for $k\geq 1$ odd. By assumption, $M$ is connected; choose a simply path between the two zeros, and a small neighborhood thereof. Now, along this path, insert in pairs zeros with Morse index $\mu +1$, $\mu + 2$, ... $\mu + (k-1)$. Now cancel the zeros in pairs but this time cancel the zero with Morse index $\mu$ with the new one with index $\mu + 1$, and so on. After all this, you end up with a vector field with two fewer zeros than you started with.

Then the result follows by induction, provided the crux in the 4th paragraph is correct.