There is a relatively simple symmetry argument that explains why there can be no "cross" polarization when a system is viewed "edge-on". The argument revolves around the fact that a binary with aligned spins and angular momentum (i.e. a binary with a well-defined, non-precessing orbital plane) satisfies a relection symmetry in the orbital plane (i.e. in standard spherical coordinates the system is invariant under the transformation $\theta\mapsto \pi-\theta$).
Consequently, the gravitational field (including gravitational waves) generated by the binary needs to satisfy this same symmetry.
Now consider what this symmetry does with a gravitational wave travelling in the orbital plane:
- A "plus" polarized wave (i.e. one with one of its axes aligned with the orbital plane, and the other perpendicular to it) is mapped to itself under this transformation.
- A "cross" polarized wave (i.e. one with both axes at 45 degree angles with the orbital plane) on the other hand ends up phase shifted by half a period.
That is, we have,
\begin{align}
h_{+} &\mapsto h_{+}\\
h_{\times} &\mapsto -h_{\times}\\
\end{align}
Consequently, requiring that the generated gravitational wave strain is invariant under reflection in the orbital plane, implies that $h_{\times}$ has to vanish in the orbital plane, i.e. when the system is view "edge-on".
This resolves the first part of the question.
The second part wonders how it can be that motion that is constrained to a plane can lead to displacement in a direction that is perpendicular to that plane. This due to the quadrupolar nature of gravitational waves. Stretching spacetime in one direction, contracts it in a perpendicular direction (and vice versa).
This behaviour should be intuitive to anybody that has ever played with play-dough. If you stretch a piece of dough, it will become thinner causing a displacement in a direction perpendicular to the direction you are pulling.
