What is the surface obtained by identifying antipodal points of $\mathbb{S}^1 \times \mathbb{S}^1$? This is perhaps a soft question.
Let $X=\mathbb{S}^1 \times \mathbb{S}^1$. Let $\mathbb{Z}_2$ act on $X$ by setting $(-1) \cdot (\theta,\psi)=(\theta+\pi,\psi+\pi)$. Consider the quotient space $X/ \mathbb{Z}_2$ which is obtained after identifying $ (\theta,\psi) \sim(\theta+\pi,\psi+\pi)$.

Is there a succinct description of $X/ \mathbb{Z}_2$ as some product or twisted/fibered product or something like that? 

Are there other "simple" descriptions of this space? Is it related to some projective space?

I feel like there should be a "right" terminology to describe it, or a way to recognize it as some familiar space, but I fail to see it.

I understand that identifying antipodal points on the $2$-torus embedded in $\mathbb{R}^3$ results in a Klein bottle- but this is not the same identification we are doing here:
Here we identify $(\theta,\psi)=(\theta+\pi,\psi+\pi)$, and in the embedded description we identify $(\theta,\psi)=(\theta+\pi,-\psi)$.
 A: There are many ways to prove this. 


*

*One, is to realize the torus as the quotient of ${\mathbb C}$ by the group of translations $\Gamma$ generated by the translations
$$
a: z\mapsto z+ 2\pi, b: z\mapsto z+2\pi i. 
$$
Lifting you involution $\tau: (\theta, \psi)\mapsto (\theta+\pi, \psi+\pi)$ yields the translation 
$$
c: z\mapsto z+ (1+i)\pi.  
$$
The group $\hat{\Gamma}\subset {\mathbb C}$ generated by $a$ and $c$ is an index 2 extension of $\Gamma$. By drawing the fundamental parallelogram of $\hat\Gamma$, you should be able to convince yourself that ${\mathbb C}/\hat\Gamma$ is diffeomorphic to the torus $T^2$. (Actually, this is a general fact that if $\Gamma < {\mathbb R}^2$ generated by translations along two linearly independent vectors then ${\mathbb R}^2/\Gamma$ is diffeomorphic to $T^2$.) 

*An alternative argument relies upon the classification of surfaces. The involution $\tau$ has no fixed points in $T^2$. Hence, $T^2\to S=T^2/\langle \tau\rangle$ is a covering map. The involution $\tau$ preserves orientation (for instance, since it is isotopic to the identity, but there are many other ways to see this). Thus, $S$ is a compact connected oriented surface and 
$$
\chi(S)=\frac{1}{2}\chi(T^2)=0.
$$
Hence, by the classification of surfaces, $S$ is diffeomorphic to $T^2$. In fact, every topological spaces covered by $T^2$ is homeomorphic to $T^2$ or the Klein bottle. 
A: The way I think about it is from the perspective of Lie theory. In each dimension, there is a unique compact abelian Lie group, namely, the torus $T^k$.
Now, $\mathbb{Z}/2\mathbb{Z}\subseteq T^2$ generated by $\langle (\pi, \pi)\rangle$ is normal (since $T^2$ is abelian, so we can form the quotient $Y:=T^2/(\mathbb{Z}/2\mathbb{Z})$).  Being the continuous homomorphic image of $T^2$, $Y$ must be a compact abelian Lie group, so it must be isomorphic to $T^2$ as a Lie group.  In particular, $Y$ is diffeomorphic to $T^2$.
$ \ $
Alternatively, you can explicitly write a diffeomorphism from $T^2$ to $Y$.  Define $f:T^2\rightarrow T^2$ by $f(\theta, \psi) = (\theta + \psi, \theta - \psi)$.
Note then that $$f(\theta + \pi, \psi + \pi) = (\theta + \psi + 2\pi, \theta - \psi + 0 \pi) = (\theta + \psi, \theta - \psi) = f(\theta, \psi),$$ so $f$ descends to a map $\overline{f}:Y\rightarrow T^2$.
One can easily prove that $f$ is surjective, that $f$ is two-to-one (in such a way as $\overline{f}$ is injective), $f$ is smooth, and that $d_p f$ has full rank everywhere.  It follows from this that $\overline{f}$ is a diffeomorphism.
