I do more graph theory, but we touch on a bit of topology there, so assuming that I understand your question, it's basically the same procedure.
Using a diagram such as this one, you paste the edges marked with the same letters together. You line up the two edges with the same letter in such a way that the arrows are pointing in the same direction. If you think about it for a few minutes, you'll see how this forms the handles for an orientable surface, or the crosscaps for a non-orientable surface.
For example, let's look at diagram 8.5.4 in the link. For those following along at home, this is a surface with two handles, so the flat representation is an octagon, with edge sequence
$$aba^{-1}b^{-1}cdc^{-1}d^{-1}$$
where the negative one means that the edge has direction reversed. So we start by pasting $a$ to $a^{-1}$, remembering to keep the arrows pointing in the same direction, which you can imagine forms a sort of cylinder with $b$ as an edge. Now you can paste $b$ to $b^{-1}$, which is a little harder to visualize but if you stare at it, you'll see that you have formed a handle. Now repeat with the other 4 edges, and you have a surface with 2 handles.
Just so this question has an answer, here's a more detailed sketch of the "standard $4n$-gon" gluing.
Label the (oriented) $1$-cells $a_{i}$ and $b_{i}$, with $i = 1, \dots, n$. Divide the boundary of the $2$-cell into a $4n$-gon. Pick an arbitrary edge, and proceed counterclockwise around the boundary, labeling successive edges $a_{1}$, $b_{1}$, $a_{1}^{-1}$, $b_{1}^{-1}$, $\dots$, $a_{n}$, $b_{n}$, $a_{n}^{-1}$, $b_{n}^{-1}$. (Some of these choices are not topologically important, just made for definiteness.) Attaching the boundary of the $2$-cell to the $1$-skeleton as indicated accomplishes the desired gluing.
Remarks:
Geometry and the Imagination by Hilbert and Cohn-Vossen contains a picture of the case $n = 3$. If memory serves, Algebraic Topology by Massey also has helpful diagrams.
It may help to draw a genus-$n$ surface and the $2n$ curves analogous to the upper left diagram in the linked post. (In that diagram, $a \mapsto a_{1}$, $b \mapsto b_{1}$, $c \mapsto a_{2}$, and $d \mapsto b_{2}$.) Then convince yourself that the product of the commutators, $\prod_{i} a_{i} b_{i} a_{i}^{-1} b_{i}^{-1}$, is homologous to a small loop around the $0$-skeleton, i.e., the vertex of the $4n$-gon. (The $2$-cell is attached to the "thickened $1$-skeleton" by filling in this loop with a disk.)
The bottommost picture in Neal's answer indicates how to make the construction inductive, i.e., how to cut a surface of genus $(n + 1)$ into surfaces of genus $n$ and genus $1$.
Best Answer
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)