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).
The definition: Let $X: U \subset {\mathbb R}^2 \to {\mathbb R}^2$ be a vector field. To define the index we assume that the zero $z \in U$ is isolated, i.e. that there is an $r>0$ so that if $0<|w-z|<r$, then $X(w) \neq 0$. Therefore,
the unit vector field $Y:=X/|X|$ is well-defined
on the punctured disc $V:= \{w: 0<|z-w|<r\}$. A map from
$S^1=\{w: |w|=1\}$ to itself is defined by $w \to Y((r/2)\cdot w)$. (Here I multiplied by $r/2$ to get $w$ into $V$, but it is a fact that multiplication by any number less than $r$ will give same degree.) The "index" is defined to be the (topological) degree of this map. (I strongly recommend Milnor's short book "Topology from the differentiable viewpoint" for a discussion of degree.)
Each self-map $f: S^1 \to S^1$ of the circle lifts to a map from
$\tilde{f}:{\mathbb R} \to {\mathbb R}$. If $\tilde{f}$ is
monotone increasing, then the cardinality of $f^{-1}(\theta)$ does not depend on $\theta$. This cardinality is the degree of $f$.
For example, the restriction of the vector field $(x,y) \mapsto (x^2-y^2,2xy)$ defines a self-map $f$ of the circle. By introducing the change of coordinates $x = \cos(\theta)$ and $y=\sin(\theta)$, we lift the map to a map $\tilde{f}:{\mathbb R} \to {\mathbb R}$. By the double angle formulas in trigonometry, we see that $\tilde{f}(\theta)= 2 \theta$.
The map $\tilde{f}$ is monotone increasing, and the number of points in $f^{-1}(\theta)$ equals $2$ regardless of which $\theta$ we choose.
Thus, the vector field $(x^2-y^2, 2xy)$ has index 2.
If the lift $\tilde{f}$ of some self-map $f$ were monotone decreasing,
then the degree of $f$ would equal the negation of the number of pre-images. For example, the vector field $(x^2-y^2, -2xy)$ has index -2.
The example of degree 2 can be understood geometrically by considering
a vector field on the sphere described as follows:
Choose a line that is tangent to the sphere at the north pole.
For each plane containing this line, the intersection of each plane
with the sphere is a circle. These circles `foliate' the sphere.
By choosing vectors tangent to the circles, one forms a
vector field on the sphere. If one looks down at the north pole,
then this vector field looks like $(x^2-y^2, 2xy)$. (A picture of
the vector field appears on page 33 of Milnor's book.)
Since the Euler characteristic of the sphere equals 2, the
Poincare-Hopf theorem implies that such a vector field has index 2
(as we understood from above).
Best Answer
Completely possible. Consider $$ F(x, y) = [x^2, 0]. $$
That's got a zero at the origin whose index is zero. (Note, too, that this field is continuous, differentiable, etc.)
The zero at the origin of the field above is not isolated, however --- it's zero everywhere on the $y$ axis. A more "typical" example is $$ G(x, y) = (x^2 + y^2) [1, 0] $$ which is nonzero everywhere except the origin, but has index zero there.