Consider this situation: There is an edge between two vertices, with three faces (maybe half-disks or half-squares, it doesn't really make a difference to topology as far as I know) going out from it, each of which meets the others only along the edge. I remove a point in the middle of the edge. I need to figure out how to prove rigorously what the homology groups of the resulting space are.
This has come up in my effort to solve a larger problem which is basically done except for this one piece. Some modeling with a piece of paper has led me to suspect it should be possible to move the space around until it resembles a wedge sum $S^1 ∨ S^1$, by something like expanding the hole made by removing the point a bit, making two of the faces coplanar and "folding up" the edge and the noncoplanar face until it's just a point, which should also make the two coplanar faces into circles or something that retracts to circles. I'm concerned my model might be misleading me, and even if it's not, I'm not sure how to clearly word an explanation of it. It seems a little strange, since it feels like I would somehow be "splitting" one hole into two holes, and that doesn't seem quite right.
The words you are missing are "homotopy equivalence". What you are saying in your second paragraph is that your space is homotopy equivalent to $S^1 \vee S^1$. This is indeed true.
But in a sense that's making things too complicated.
A more straightforward statement with the same outcome but which is easier to prove, is that the union of the six non-punctured edges is a deformation retract of your space.
Once you've proved that, you then notice that union of those six edges is homeomorphic to a finite 1-complex with two vertices, say $V,W$, and with three edges, say $E,F,G$, such that each of $E,F,G$ has one endpoint at $V$ and the other at $W$. You can then proceed to compute the homology groups of that 1-complex. Or, instead, you could prove that that 1-complex is homotopy equivalent to $S^1 \vee S^1$, ending at last where you expected to end.