From the Leray-Serre spectral sequence for a covering map $Y\to X$, which is a fibration with discrete fibers, we get an isomorphism $H^p(Y)\cong H^p(X,\mathcal H^0)$, where $\mathcal H^0$ denotes the local system of coefficients which at each point of $X$ has group equal to $H^0(p^{-1}(x))$.
If the covering is of $n$ sheets, then $\mathcal H^0$ is locally $\mathbb Z^n$, with the fundamental group of $X$ acting by permutation of the standard basis according to the monodromy permutation representation.
Now the local system $\mathcal H^0$ corresponds to a sheaf $\mathcal F$ on $X$, and for sensible $X$ (paracompact, say), one can compute singular cohomology with coefficients in the local system as sheaf cohomology with coefficients on the sheaf $\mathcal F$. If $X$ has a good finite cover $\mathcal U$ (in the sense of the book of Bott-Tu) then one can compute sheaf cohomoogy $H^p(X,\mathcal H^0)$ as the Cech cohomology $H^p(\mathcal U,\mathcal H^0)$. Looking at the complex which computes this by definition, we see that the Euler characteristic of $H^p(\mathcal U,\mathcal H^0)$, and therefore of $H^\bullet(X,\mathcal H^0)$, is $n$ times that of $H^\bullet(X,\mathbb Z)$. Notice that the existence of good covers implies being of bounded finite type, as you say (but I think it even implies that the space of of the homotopy type of a CW-complex, namely the nerve of the good covering... so all this might not get us much)
The fact that the Euler characteristic of a sensible space with coefficients on a local system of coefficients which locally looks like $\mathbb Z^n$ is $n$ times that of the space should be written down somewhere, but I cannot find it now.
There is this answer by Matt but he does not give a reference.
Basically, this is correct, but $V$ should map to $\mathbb R^3$ (but this makes sense for surfaces not sitting in $\mathbb R^3$ as well) and to make sense of degree you need to look at a local-coordinate formulation of $(V\circ f)/\|V\circ f\|\colon S^1\to S^1$.
The converse result is quite deep and holds in all dimensions. I recommend you look at Milnor's Topology from a Differentiable Viewpoint and Guillemin and Pollack's Differential Topology. Morris Hirsch's book by the same title is quite a bit more advanced.
Of course, it's quite easy to give a non-vanishing vector field on the torus. Just comb the hairs on each circle. If you pick the right set of circles, this vector field should descend to a well-defined vector field on the Klein bottle (remember the torus is a two-fold covering space of the Klein bottle).
Best Answer
$\newcommand{\Z}{\mathbb{Z}}$The Klein bottle has the following integral homology groups: $$H_*(K; \Z) = (\Z, \Z \oplus \Z/2\Z, 0, 0, \dots).$$ This gives an Euler characteristic $\chi_\Z = 1-1 = 0$. Over $\Z/2\Z$, the universal coefficient theorem yields: $$H_*(K; \Z/2\Z) = (\Z/2\Z, \Z/2\Z^2, \Z/2\Z, 0, 0, \dots)$$ and so the Euler characteristic is $\chi_{\Z/2\Z} = 1-2+1 = 0$. This agrees with the previous computation.
In general, suppose $X$ is a space with finite integral homology (i.e. a finite number of nonzero homology groups, and these groups are all finitely generated), for example a finite CW-complex. Then its Euler characteristic is well-defined. Let $n$ be the top dimension of nonvanishing homology.