Fundamental group of $T/\mathbb Z_2\setminus\{\text{singular points}\}$ Let $T=S^1\times S^1$. There is a $\mathbb Z_2$-action on $T$ defined by $x\sim -x$ (considering $T$ as a quotient of $\mathbb R^2$).
The quotient $X:=T/\mathbb Z_2$this has four singular points, say $x_1,...,x_4$ (corresponding to the four half lattice points in  $\mathbb R^2$). I want to calculate the fundamental group of $X\setminus\{x_1,...,x_4\}$.
So far I could think of the following:
Consider $Y:=T\setminus\{x_1,...,x_4\}$ (abuse of notation). Then $Y\tilde = S^1\vee S^1\vee S^1\vee S^1\vee S^1.$ Now $X=Y/\mathbb Z_2$. So we need to check how $\mathbb Z_2$ acts on the $S^1$-summands.
From these we can choose three summands to be small circles around a singularity. The $\mathbb Z_2$-action on them is then simply the $\mathbb Z_2$-action on a circle with center $0$ in $\mathbb R^2$.
So for them $S^1/\mathbb Z_2=\mathbb RP^1=S^1$. 
But what about the other two summands? They seem to be glued together in a way, such that the $\mathbb Z_2$ action acts more complicatedly.
Update: Thanks to the comment by Mike Miller, I was able to look up the construction of the "pillowcase". SO this is basically answered.
However, I would be interested in generalisations of this in higher dimensions. For example $T^4/\mathbb Z_2$. 

 A: You may think of $T^2 = \Bbb R^2/\Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $\Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $\Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) \sim (1,t)$ and $(t,0) \sim (t,1)$. 
The action of negation on the unit square in $\Bbb R^2/\Bbb Z^2$ is to send $$(x,y) \mapsto (-x, -y) \sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(\frac 12, \frac 12)$. 
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] \times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) \sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there.  We now have a cylinder.
We also know from the reflection relation that $(1/2, t) \sim (1/2, 1-t)$ and $(0,t) \sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] \times [0,1]^{n-1}$.
