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.

For $\Sigma$ of higher genus, the model geometry will be $\mathbb{H}^2 \times \mathbb{R}$, and the Seifert fibering will be unique.

I think that's right. Given your hypotheses, such a manifold should be constructed by taking a finite dihedral subgroup of the homeomorphism group of a surface $\Sigma$ with two (fixed-point free) involution generators $\alpha, \beta$, 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 by projecting to the second coordinate (so is a semi-bundle) and is Seifert-fibered by projecting to the (quotient of) the first coordinate. 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. However the base orbifold should be the same for each of the three fibrations.

For $\Sigma$ of higher genus, the model geometry will be $\mathbb{H}^2 \times \mathbb{R}$, and the Seifert fibering will be unique.

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.

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.

For $\Sigma$ of higher genus, the model geometry will be $\mathbb{H}^2 \times \mathbb{R}$, and the Seifert fibering will be unique.