Here's an idea; I'm not sure if all of the steps are possible, but I think it should work.

Take a simplicial complex of dimension $n$ (all simplices are of dimension $\leq n$). "Thicken" the simplicial complex to get a new topological space containing the original simplicial complex as a deformation retract, such that there is an open neighborhood $U_\sigma$, homeomorphic to an $n$-ball, of each simplex $\sigma$ which deformation retracts to that simplex. Do this so that the intersections $U_\sigma \cap U_{\sigma'}$ are equal to $U_{\sigma \cap \sigma'}$. Now take the open cover $\{ U_{\sigma} \}$; then the Cech complex associated to this open cover should be the "same" as the complex that computes the simplicial (co)homology of the simplicial complex. "Same" here probably means "canonically homotopy equivalent".

For example, if you think of a circle as being the simplicial complex given by a triangle with 3 edges and 3 vertices, then this essentially matches up with your example (2).

This is essentially the same as Mariano's explanation, but sort of in the converse direction.

This sort of thing might be covered in some older books, like Munkres' "Elements of Algebraic Topology" or Spanier's "Algebraic Topology". Cech cohomology doesn't seem to really be covered in most contemporary algebraic topology books; I guess it's just not very useful for what people are interested in these days.