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The space is the mapping torus of the antipodal map $A:S^2\to S^2$. Its mapping degree is $(-1{)}^3=-1$, so transporting a fiber orientation once around the base circle reverses it. $X$ is nonorientable.

On integral fiber homology, $A_{\ast }$ is the identity in degree zero and minus the identity in degree two. The Wang sequence, whose relevant map is $1-A_{\ast }$, therefore givesThe universal coefficient theorem for cohomology shifts the torsion contribution into degree three, because ${\mathrm{Ext}}_{\mathbb{Z}}(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$. HenceLet $t\in H^1(X;\mathbb{Z})$ be pulled back from the base circle, and $u$ the nonzero degree-three class. The square of the circle class vanishes, so $t^2=0$. All other products of positive-degree classes land above dimension three and vanish. This gives the cohomology ring of the antipodal two-sphere mapping torusModulo two, $A^{\ast }$ is the identity on fiber cohomology. The cohomological Wang sequence shows that restriction $H^2(X;{\mathbb{F}}_2)\to H^2(S^2;{\mathbb{F}}_2)$ is onto; choose $y$ mapping to its generator. The classes $1,y$ restrict to a fiber basis, so the Leray-Hirsch theorem makes the total cohomology free over $H^{\ast }(S^1;{\mathbb{F}}_2)$ on those two classes. Write $\bar{t}$ for the mod-two base class. The basis is $1,\bar{t},y,\bar{t}y$, and in particular $\bar{t}y\ne 0$. Since ${\bar{t}}^2=0$ by pullback and $y^2=0$ by dimension, the mod-two cohomology ring of the antipodal two-sphere mapping torus isThe argument of Question 3 works over any field, so its use here over ${\mathbb{F}}_2$ is justified. The ring agrees with that of $S^1\times S^2$; a cohomology ring alone does not determine orientability.