1) Near a point of $S^2$ the picture looks like $\mathbb{C}$ sitting inside, naturally $\mathbb{C}^2$ (because $\mathbb{C}P^2\cong(\mathbb{C}^3\setminus{0})/\mathbb{C}^{\times}$ and $S^2=\mathbb{C}P^1$ is the image of a hyperplane in $\mathbb{C}^3$).
2) Think first about $S^2$: one attaches a disk to a point, contracting the circle on the disk's boundary. Now, to make $\mathbb{C}P^2$ one takes 4-disk and do pretty much the same but not for one circle but for all, well, fibers of the Hopf fibration at once — or, in other words, for all points of $S^2$ at once. So, since there were no singularities after gluing $S^2$, there will be no singularities here either.
(Not sure if it's an answer, but only hope it helps.)
Fortunately, the one you understand can be readily seen in the cell-complex construction.
So, take a rectangle, identify opposite sides. Now, draw a picture of a torus, and draw the rectangle on it. This is very important that you can do this. What does the rectangle look like on the torus? It likes like a sort of figure-8 (sort of). All four corners are the same point. Two pairs of opposite sides are associated, so we get only 2 edges, not 4. Good.
This is how the cell-structure comes, too. Take one point (0-cell). Take 2 1-cells (each loop in the figure 8). And take 1 2-cell. But how do we attach our 2 cell? Well, a 2-cell is just a square. So on the original rectangle that you drew and understood, why don't you just take that to be your 2-cell? Then the attaching maps are precisely those implied by your drawing.
So the cell-structure and the rectangle are, in fact, the exact same. In fact, when I give cell structures for genus-g surfaces, I give them in that fashion.
It all comes down to (in my opinion) finding that figure 8 on the torus itself, to understand what that rectangle is. If this doesn't make sense, comment, and I'll upload an image.
Here, we see the figure 8 and the two loops. All four corners are the same point. Their intersection is the 0-cell, the red and the blue are each 1-cells, and the surface is a single 2-cell is attached with the following attaching map (where a is the red side, b is the blue)
Best Answer
Geometrically, the pictures for both cases you provide are correct and are the same.
However, $abab$ would mean that while you traverse the boundary of the square, $a$ and $b$ are pointing towards the direction of your movement both times you meet them. This is not the case: starting from the lower left corner and moving clockwise, we go first along $a$ and $b$, but then meet $a$ pointing backwards relative to our movement. So $aba^{-1}b^{-1}$ is the correct way to write it.