= Solution
Use <linearization> around <Minkowski spacetime> after constant rescaling of $x,y$: set $A=1+\varepsilon a(u)$, $B=1+\varepsilon b(u)$, $u=t-z$. To first order,
$$
g=\eta+2\varepsilon a(u)\,dx^2+2\varepsilon b(u)\,dy^2+O(\varepsilon^2),\qquad a''+b''=0.
$$
Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The <affine transverse trace in a diagonal linearized plane wave is pure gauge>: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order <linearized coordinate gauge transformation> $h_{\mu\nu}\mapsto h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$, with
$$
\xi_x=\frac{\varepsilon s x}{2},\quad \xi_y=\frac{\varepsilon s y}{2},\quad \xi_t=-\frac{\varepsilon s'(x^2+y^2)}4,\quad \xi_z=\frac{\varepsilon s'(x^2+y^2)}4.
$$
The $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s''=0$. The transverse perturbation becomes
$$
\boxed{h_{xx}=\varepsilon h(t-z),\qquad h_{yy}=-\varepsilon h(t-z),\qquad h_{xy}=0,}
$$
with all time and longitudinal components zero. It is a <transverse-traceless gauge> plane <gravitational wave> travelling in the positive $z$-direction, with <plus polarization> and no <cross polarization> in these transverse axes.
For freely falling nearby particles, <geodesic deviation> gives opposite transverse tidal accelerations, since $R_{txtx}=-\varepsilon a''$ and $R_{tyty}=-\varepsilon b''=+\varepsilon a''$ in vacuum. Thus the transverse <geodesic deviation> <eigenvalues> have opposite signs; the stretching and compressing roles interchange when the <curvature> profile changes sign. If $h''=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.
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