What is the surface obtained by identifying antipodal points of $\mathbb{S}^1 \times \mathbb{S}^1$?

There are many ways to prove this.

1. 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 $T^2$.)

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.

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 $T^2$.

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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.