Let $p:S^2\to P^2$ be the double covering of the real projective plane. Let $g:P^2 \to P^2$ be a map such that its induced homomorphism on fundamental group is not trivial.
I'd like to show that there exists $T:S^2\to S^2$ covering $g$ with $T(-x)=-T(x)$.
Given $x\in S^2$ take $\bar x=p(x)\in P^2$ and $\bar y=g(\bar x)\in P^2$. It is clear that the fiber over $\bar y$ has two points $y_1,y_2\in S^2$.
I guess that $T(x)$ should be one of them. Also I guess that the action of $\pi_1(P^2)$ should be relevant to choose between $y_1,y_2$. But I'm not sure how.
I tried to take $\alpha\in\pi_1(P^2,\bar x)$ a generator and considered its image $\beta=g_*(\alpha)\in\pi_1(P^2,\bar y)$. Then I tried to take a lifting. But what should be the criteria to choose the necessary points?
You're right that there exists such a $T$.
The map $(g\circ p)_*:\pi_1(S^2)\rightarrow \pi_1(P^2)$ has trivial image because $S^2$ is simply connected. It follows that $g\circ p$ lifts to a map $S^2\rightarrow S^2$ because $S^2$ is $P^2$'s universal cover. (This is an application of the "lifting theorem" of covering space theory. See the accepted answer to this question.) This is the desired $T$.
To check that $T(-x)=-T(x)$, take a path from $x$ to $-x$ in $S^2$, and push it down to $P^2$ via $p$; it becomes a generator for $\pi_1(P^2,\bar x)$ where $\bar x = p(x)$. Push it forward to $P^2$ via $g$; it remains non-nullhomotopic because of your assumption about $g_*$.
On the other hand, by first pushing the path forward to $S^2$ along $T$ and then down to $P^2$ via $p$, we see that $T(x)$ and $T(-x)$ must be distinct; otherwise the loop would end up nullhomotopic. But if they are distinct, they are opposites, because they both map to $g(\bar x)$ via $p$.