As Sam Nead points out, it's not correct that every Seifert-fibred space is finitely covered by a product. The obstruction to this is the Euler number of the bundle. I strongly recommend Peter Scott's article 'The geometries of 3-manifolds' if you want to find out more about Seifert-fibred spaces. A scanned copy is available on his web page.
In the case when the base orbifold $O$ is torsion-free (and this is something that you always can assume by passing to a finite-sheeted cover---the point is that we only need to consider 'good' orbifolds, see Scott for details), the Euler number can be thought of as the obstruction to the short exact sequence
$1\to \mathbb{Z}\to\pi_1(M)\to\pi_1(O)\to 1$
splitting. Of course, this always happens when $\pi_1(O)$ is a free group, ie when $O$ is a surface with boundary.
This is why Hempel says you can assume that a finite-sheeted cover is a product---because he really cares about the case in which the JSJ decomposition is non-trivial. In this case, the Seifert-fibred pieces all have boundary, and so do indeed have a finite-sheeted cover that's a product.
Finally, it is true that you can prove residual finiteness of Seifert-fibred manifolds just from the short exact sequence. But care is needed! Deligne gave a famous example of a central extension of a lattice in a Lie group which is not itself residually finite.
For this reason, the proof is quite fiddly. The idea is to reduce to the case of a finite central extension of a surface group. You can in fact prove that any finite central extension of any residually finite one-relator group is residually finite, using Magnus's theorem that free groups are residually torsion-free nilpotent.
For full details, you could look for instance at this paper of Martino, in which he proves the stronger fact that Seifert-fibred 3-manifold groups are conjugacy separable.