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.
Best Answer
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.