Realising the non-trivial orientable $S^2$-bundle over $T^2$ as a quotient of $S^2\times T^2$ Using the fact that $\operatorname{Diff}^+(S^2)$ deformation retracts onto $SO(3)$, one can show that over any connected, closed, smooth surface, there are two orientable $S^2$-bundles, the trivial one and a non-trivial one. Moreover, the two bundles can be distinguished by their second Stiefel-Whitney class.
Let $X$ be the total space of the non-trivial orientable $S^2$-bundle over $T^2$. If we pullback $X \to T^2$ by any double covering $T^2 \to T^2$, we obtain the trivial $S^2$-bundle $S^2\times T^2 \to T^2$. It follows that $X$ is double covered by $S^2\times T^2$.

What is the free $\mathbb{Z}_2$ action on $S^2\times T^2$ which has quotient $X$?

It would be enough to realise the total space of an orientable rank two bundle $E \to T^2$ with $w_2(E) \neq 0$ as a $\mathbb{Z}_2$ quotient of $T^2\times\mathbb{R}^2$.
Note that $X$ can be realised as the sphere bundle of $\gamma_1\oplus\gamma_2\oplus(\gamma_1\otimes\gamma_2)$ where $\gamma_i$ is the pullback of the non-trivial line bundle on $S^1$ by projection onto the $i^{\text{th}}$ factor. It is easy to see pulling back by $(z, w) \mapsto (z^2, w)$ (or $(z, w) \to (z, w^2)$), the bundle becomes trivial (as it must). I was hoping these explicit descriptions would help, but so far I have been unsuccessful.
 A: For flat $SO(3)$-bundles over a (connected oriented) surface $\Sigma$ the 2nd SW-class is the obstruction to lifting the associated representation $\rho: \pi_1(\Sigma)\to SO(3)$ to a representation $\tilde\rho: \pi_1(\Sigma)\to SU(2)$.
Now, if $\Sigma=T^2$ with generators of the fundamental group $a, b$, the standard example of a non-liftable representation $\rho: \pi_1(\Sigma)\to SO(3)$ is the one whose image is $Z_2\times Z_2$ (sending the generators $a, b$ to the generators of the direct factors). If one thinks of $S^2$ as the Riemann sphere and $SO(3)$ as acting by linear-fractional transformations, WLOG, the representation $\rho$ is described as
$$
\rho(a): z\mapsto -z, \rho(b): z\mapsto z^{-1}. 
$$
Consider, therefore, the product $Y=T^2\times S^2$ and $Z_2\times Z_2$ (with generators $\alpha, \beta$) acting on $Y$ via:
$$
\alpha\cdot (u, v) \times z= (-u, v) \times (-z),
$$
$$
\beta\cdot (u, v) \times z= (u, -v) \times (z^{-1}).
$$
(Here $u, v\in S^1$.) The quotient space of this action is $X$, the total space of  your oriented nontrivial $S^2$-bundle over $T^2$.
One can (in principle) extract from this description the one for a 2-fold covering space
$$
T^2\times S^2\to X.
$$
Namely, first divide $Y$ by the action of the first factor of $Z_2\times Z_2$. Then, one has to trivialize the quotient bundle $W=Y/Z_2\to T^2$.  This can be done by observing that the map $Y\to S^2$, $(u,v)\times z\mapsto z^2$ is $\alpha$-invariant, hence, descends to a map $W\to S^2$ and, therefore, can be used for a trivialization. Then using the resulting (biholomorphic) diffeomorphism $h: W\to T^2\times S^2$ one obtains the (holomorphic) action of the involution $\beta$ on $T^2\times S^2$  by conjugation via $h$. From this, one concludes for instance that there exists a degree 2 holomorphic covering map $T^2\times S^2\to X$. Writing an explicit holomorphic action of $\beta$ on $T^2\times S^2$ from this is doable but feels unpleasant. If you are sufficiently motivated, I am sure you can write down an explicit formula.
A: Just for kicks, I wanted to explicitly write down the $\mathbb{Z}_2$ involution on $Y:=T^2\times S^2$, following the description in Moishe's answer.  For concreteness, I'm going to view $T^2\subseteq \mathbb{C}^2$ and $S^2$ as the unit sphere in $\mathbb{C}\oplus \mathbb{R}$.  Thus, a point in $S^2$ has the form $(x+iy, z)$ with $(x,y,z)\in \mathbb{R}^3$ and $x^2 + y^2 + z^2 = 1$.

The involution on $T^2\times S^2$ with quotient $X$ is given by $$((u,v),(x+iy,z))\mapsto ((u,-v), (-u\overline{(x+iy)}, -z)).$$

The rest of this post is devoted to justifying this formula.
We begin by converting Moishe's $\alpha$ and $\beta$ into one more useful for our description of $S^2$.  Under the usual identification of $\mathbb{C}\cup \{\infty\}$ with $S^2$, the map $\alpha$ of Moishe's answer has the form $(x+iy,z) \mapsto (-(x+iy),z)$ on the $S^2$ factor of $Y$.  Similarly, the map $\beta$ of Moishe's answer has the form $(x+iy,z)\mapsto (-x+iy, -z) = (-(\overline{x+iy}), -z)$.
As Moishe indicates, the space $Y/\langle \alpha\rangle$ is diffeomorphic to $Y$.  But we need an explicit diffeomorphism.
Proposition:  The map $\phi:Y\rightarrow Y$ given by $\phi((u,v),(x+iy,z)) = ((u^2,v), (u(x+iy), z) $ descends to a diffeomorphism $\Phi:Y/\langle\alpha\rangle \rightarrow Y$.
I'll leave the proof to the interested reader.  Because we'll need it later, let me note that $\Phi^{-1}$ is the map defined by $$\Phi^{-1}((u,v), (x+iy, z)) = [((\sqrt{u},v), \overline{\sqrt{u}}(x+iy), z)]$$ where $\sqrt{u}$ means any particular fixed choice of square root of $u$.  That is, it doesn't matter which of the two roots you use for $\sqrt{u}$, but you have to use the same one both places "$\sqrt{u}\,$" appears.
Now, we define an action of $\mathbb{Z}_2 = \langle \gamma\rangle$ on $Y$ by declaring $\Phi$ to be equivariant, where $\mathbb{Z}_2$ acts on $Y/\langle \alpha \rangle$ via $\beta$.
That is, $\gamma \ast p$ for $p\in Y$ is defined to be $\Phi(\beta \ast \Phi^{-1}(p))$.
Carrying out this computation, we find that $$\gamma\ast((u,v),(x+iy),z) = ((u,-v), -u(\overline{x+iy}), -z)).$$
As a sanity check, we should have $\gamma^2 = Id$. And, indeed, \begin{align*} \gamma^2((u,v),((x+iy),z)) &= \gamma\ast((u,-v), -u(\overline{x+iy}), -z))\\ &= ((u,v), u(\overline{u(\overline{x+iy})}), z))\\ &= ((u,v),(u\overline{u}(x+iy), z))\\ &= ((u,v),(x+iy,z)),\end{align*} as claimed.
