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)
Take the dodecagon at the origin with one pair of edges intersecting the $y$-axis (call them the top and bottom faces) and one pair intersecting the $x$-axis. Cut the polygon along the $x$ axis, and un-identify the left and right faces. This gives two octagons, each with an opposing pair of unmatched edges. Identify these new edges, and you should have what you're looking for.
Best Answer
Imagine a closed chain of tori (that is, torus $i$ is connected-summed to tori $i+1$ and $i-1$ mod $2k.$) That is a (cyclic) $k$-fold cover of a double torus (surface of genus $2.$).