Suppose you have an $n$-gon with all vertices identified, and all edges identified. I think the optimal way to compute the homology groups would be to view this as a cell complex consisting of a single $0$-cell, $1$-cell, and $2$-cell. Clearly the cellular maps $d_0,d_1\equiv 0$, and the degree of $d_2$ could be computed by summing up $\pm 1$ depending on the orientations of the edges.
For curiosity's sake, how could you go about this with simplicial homology? If $n>3$, I think you'd have to add another $n-3$ diagonals to break up the $n$-gon into a bunch of triangles.
Would all these new diagonals would be completely new edges? How would we know which way to orient them? I think $d_1$ and $d_0$, would still be easily seen to be $0$, but would computing $d_2$ require us to compute the boundary of all $n-1$-triangles in the $\Delta$-complex, and somehow determine the quotient of $\mathbb{Z}$ by whatever the generated image is?
Yes. They do not seem to be part of your description of an $n$-gon. Notice you will need more vertices, too, if you want the resulting object to be a simplicial complex. (Your construction is not well-defined for exactly this reason. How do you add new edges? Where? How are you ensuring that the resulting object is a simplicial complex?)
You get to choose the orientation, and the homology groups are independent (up to isomorphism) of your choice of orientation. The resulting generators of homology, however, do depend on these choices.
The map $d_0$ is clearly still $0$ (unless you are computing reduced homology). Because you added new vertices (to ensure that the object is indeed a simplicial complex), $d_1$ is NOT zero.