We know that $\chi(M) = 2 = \sum_{x_0 \in \text{Sing}(X)}I(x_o,X)$, where $X$ is any tangent vector and $x_0$ is a singularity. I mean, the sum is over all singularities of $X$, an arbitrary tangent vector field.
Then, once this sum is equal $2$ for every vector field $X$, all of these vectors has at least a point where vanishes.
Is this right?
I think it makes more sense to say 'zero' rather than 'singularity', but yes. If $X$ is a vector field on $M$ which has isolated singularities, and $\chi(M) \neq 0$, then $X$ must have at least one singularity. Otherwise, the sum over singularities is empty and hence zero which is impossible as $\chi(M) \neq 0$.