i Want to do an exercise out of the book three dimensional Geometry and topology by thursten.
Which states Any isometry of $\mathbb{S}^3$ that has no fixed points is orientation-preserving. Inparticular every elliptic three-manifold (a quotient $\mathbb{S^3}/G,\;\;\;G \subset O(4)$) is orientable.
The first statement isn't to important for my application, but I would need the second statement and I have no clue how to even start proving it.
Here's a very useful general (and intentionally vague) principle to have in mind when dealing with quotients: if the action of $G$ on $M$ preserves some structure on $M$, then that structure should descend to the quotient $M/G$. That's super vague, so here's a more concrete claim: if $G$ is finite (so that the quotient map is a local diffeomorphism), then any tensor field on $M$ that is invariant under the action of $G$ descends to $M/G$ (i.e. is the lift of some tensor field on $M/G$).
In this case, the fact that the group action preserves orientation implies that the orientation on $M$ induces an orientation on $M/G$. The precise proof of this will depend on exactly how you're defining orientation. If you use volume forms, you can just average the volume form by the action of $G$ (since it's finite, right?) to get a nonvanishing $G$-invariant volume form on $M,$ which then descends to a volume form on $M/G.$
For $S^3$: Assuming that your group action is free, all non-identity $g \in G$ have no fixed points and thus (by the first statement) are orientation-preserving; and the identity is obviously orientation-preserving.