I think that's right. Given your hypotheses, such a manifold should be constructed by taking a finite dihedral subgroup of the mapping class group of a surface $\Sigma$, say with two (fixed-point free) generators realized by orientation-reversing involution homeomorphisms $\alpha, \beta$, and creating a manifold by taking $M\cong \Sigma\times[0,1]/\{ (x,0)\sim (\alpha(x), 0), (x,1)\sim (\beta(x),1)\}$. They need to be orientation-reversing so that $M$ is orientable, and by construction this fibers over a mirrored interval (so is a semi-bundle) and is Seifert-fibered. Then the base of the Seifert-fibration will be homeomorphic to $\Sigma / \langle \alpha, \beta \rangle$, a non-orientable 2-orbifold.
As an example, if $\Sigma \cong S^2$, then one should have the group $\mathbb{Z}/2\mathbb{Z}$ realized by $\alpha=\beta$ the antipodal involution of $S^2$. Then the manifold will be homeomorphic to $\mathbb{RP}^2\#\mathbb{RP}^2$ fibering over $\mathbb{RP}^2$. Notice that the fibration will be non-orientable to match the non-orientability of the base.
For $\Sigma \cong T^2$, I think that there are 3 possible conjugacy classes of finite dihedral subgroups of the mapping class group generated by non-orientable involutions. However, I think only two of these are realized by a manifold, and without a unique Seifert-fibering. One is the orientable circle bundle over the Klein bottle. The other has base a projective plane with two cone points of order 2 ($22\times$ in orbifold notation). One may determine this by the classification of wallpaper groups. The desired manifold is the unit tangent bundle to these orbifolds, with Seifert-fibering given by the unit tangent circles. But one may also "pull back" fiberings from the base orbifold to get Seifert fiberings of the manifold.