what is the nature of a ball that goes over a "corner" of the real projective plane? I'm make a little computer program to help me understand different 2d topological spaces, (such as torus and mobius band). 
I'm having issues with drawing balls that go over a corner of the real projective plane. By "corner" I am referring to a rectangle that is mapped to the projective plane as shown in the wolfram link. 
The images below were drawn with my program. It shows the boundary of a ball drawn around point p. The top image make sense to me (the red, blue, orange, and purple dots show points that are identified with one another). The bottom picture shows my programs interpretation of when the ball go over the corner, and you can see that the circles in the bottom right are overlapping. 

Does the ball around p, in the bottom case, contain all points in either circle at the bottom right (i.e. their union)? Is my program not drawing the balls correctly? 
 A: When a rectangle's edges are identified to make a torus or Klein bottle (each of which admits a flat metric), the corner geometry is isometric: The four right angles add up to $2\pi$.
Contrast with the situation of the projective plane: At the corners of the Euclidean rectangle, the model does not conformally represent a neighborhood. Instead, there are cone points, one for each pair of diagonally opposite vertices. Each cone point has incident angle $\pi$, from the two right angles.
In your top picture, the disk does not enclose the cone point at the northwest/southeast corner, but you can see the disk's boundary "puckering" near the southeast corner.
In the bottom picture, the disk contains a cone point, and the boundary circle winds twice around it. (The circle's unit tangent field has a total turning of $2\pi$, but there's only an intrinsic angle of $\pi$ incident at the cone point.)
If it's feasible to represent the projective plane by using antipodal boundary identifications on a disk (which eliminates the cone points), you'll get boundary behavior that better aligns with intuition.

Edit: There's a pleasant cross-cap model of the projective plane that can be made in about one minute from a rectangular sheet of paper (and taped closed as desired) to illustrate physically what's happening with the disk in the original question.

Fold the paper in half along the vertical midline. Unfold, and repeat along the horizontal midline. The sheet should now have two perpendicular valley folds running diametrically across.
Slit the paper from the midpoint of the top side to the center (the bold segment). Fold quarter I down onto II, then fold II rightward onto III, then fold III upward onto IV. All four quarters lie over IV in a four-ply arrangement. Identify (e.g., tape closed) the right and top edges of the top two plies (I and III), and the right and top edges of the bottom two plies (II and IV). The slit, running along the left edge, must be mentally re-joined; this (together with the actual segment joining II and III) is the line of self-intersection of the cross cap.
When I and II are folded rightward onto III, the top edge of I and the bottom edge of III are brought together with opposite orientation, and the left edge of I and the right of III are brought together with opposite orientation. Similarly, when I, II, and III are folded upward onto IV, the free edges of II and IV are brought together with the correct orientations to make a projective plane.
The finished (taped) model can be opened up into a double-sheeted cone (like a paper hat) whose vertex is the center of the original sheet. The "cone points" are the corners of the original sheet. At each, only two right angles' worth of paper touch, so each point has angular defect $\pi$. At every other point (including the center of the sheet), a small neighborhood has incident angle $2\pi$. 
