= Solution
Put the rigid lid at $z=0$, the interface at $z=-h(x,y,t)$ and $\rho_1=\rho_0-\Delta\rho$. Assume <Hydrostatic equilibrium>[hydrostatic vertical balance] and a resting deep layer. Its <pressure> is $p_2=C-\rho_0gz$, with horizontally uniform $C$. <Pressure> continuity at the interface then fixes the upper-layer <pressure>:
$$
p_1=C+\rho_0gh-\rho_1g(z+h)
=C-\rho_1gz+\Delta\rho\,gh.
$$
Hence its <Gradient>[horizontal pressure gradient] at fixed $z$ is $\nabla_hp_1=\Delta\rho\,g\nabla_hh$, independent of depth. Taking the horizontal momentum equation and a vertically uniform upper-layer velocity gives
$$
\boxed{\frac{D\mathbf u}{Dt}+f\mathbf k\times\mathbf u+g'\nabla_hh=0,
\qquad g'=g\frac{\Delta\rho}{\rho_1}\simeq g\frac{\Delta\rho}{\rho_0}.}
$$
The exact <density> denominator for upper-layer acceleration is $\rho_1$; the familiar $\rho_0$ denominator is the small-density-contrast <Boussinesq approximation>. The restoring <pressure> is weaker than for a free surface because only the <density> difference contributes.
The horizontal shallow-layer equations use the vertical component of planetary rotation: $f=2\Omega\sin\varphi$. Thus the source's <vector> notation is interpreted as the projected <vector> $f\mathbf k$ after the <traditional approximation>. Taking the full magnitude $2\Omega$ as a midlatitude <Coriolis parameter> would discard the <latitude> dependence. Together with <Volume conservation>[layer-volume conservation], the model has $\partial_th+\nabla_h\cdot(h\mathbf u)=0$.
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