Hole in the topology of geometrical representation of covectors One can make geometrical models of vectors and covectors like this:

The arrow for the vector is presumably familiar to most people. The representation of the covector involves two parallel lines, with an orientation specified. If you play with this representation, you'll find that there is only one sensible way to define addition of these figures. This representation is introduced, for example, in Burke's Div, Grad, and Curl Are Dead. In the model of covectors, a smaller spacing between the lines represents a scaling up of the covector.
All of this makes sense to me, but I'm bothered by the following blemish. Because of the way that scaling is represented for covectors, we have to rule out the degenerate case where the lines coincide (it would represent a vector with infinite magnitude), and we also have no representation for the zero vector (the lines would have to be infinitely far apart). So it seems that this is a representation of the covector plane with a hole in it, which is topologically a cylinder.
Is there any more general way of understanding what's going on with the missing point? Can this representation be cleaned up to eliminate the hole, e.g., by adjoining an idealized point for zero (seems ugly)? Does this have something to do with projective geometry?
 A: This is somewhere between a comment and an answer. I think the best way to think about the covector representation is that the orientation of the lines gives "direction" of the covector (or more specifically if we think of a covector as something of the form $\langle\cdot,w\rangle$ where $\langle\cdot,\cdot\rangle$ is the inner product and $w$ is a vector, then $w$ will be orthogonal to the lines (or rather hyperplanes) represented. 
This is the straightforward part and the distance between the lines seems the less intuitive part. I think that the best way to think about the distance between the parallel lines is to think of them as analogous to a scale (or legend) on a map. 
If we let $\phi=\langle\cdot,w\rangle$ be out vector and $v$ be our vector then you can think of $\phi(v)$ being the length of $v$ that lies in the direction of $w$. Here is where the picture you posted above is quite instructive. If I were to ask you the length of the vector on the right in the direction of the covector on the right (which you specify the direction of) you couldn't give me a numerical answer because there is no numerical scale to judge size. This is exactly what the space between the lines represents. 
The idea is that you project $v$ onto $w$ and that will give you a vector and the unit of length is the width of the lines, so that if the resulting projection was twice as long as the distance between the lines then value of $phi(v)$ would be 2.
Now you can see that the closer together the parallel lines for the covector are then the larger value of $\phi(v)$ will be. 
As far as your last two questions, this representation doesn't translate well to a zero covector and I don't think that this really has much to do with projective geometry aside from the fact that the space of covectors represented in this ways seems to be a double cover of the canonical line bundle over real projective space. 
