My answer is in three parts: (1) why the horizontal strain in the seafloor is proportional to pressure, (2) how the strain the cable relates to the strain the seafloor, and (3) why the pressure gradient might fit your data well.
(1) I just re-derived this to make some synthetics the other week, but I think it’s the same as Wayne Crawford’s thesis.
For a homogeneous elastic half-space in 2D, we can write the displacement u in P (\phi) and S (\psi) potentials as u_x = \partial_x\phi - \partial_z\psi and u_z = \partial_z\phi + \partial_x\psi. When inserted into the elastic wave equation, this “Helmholtz decomposition” yields two nominally independent wave equations for P and S waves which are only coupled by the boundary conditions:
\partial_{tt} \phi = \alpha^2 \left( \partial_{xx} + \partial_{zz} \right)\phi and \partial_{tt} \psi = \alpha^2 \left( \partial_{xx} + \partial_{zz} \right)\psi
The boundary condition at the seafloor enforced by a plane harmonic surface wave is that the normal stress at the seafloor must equal the pressure at the bottom \sigma_{zz}(z=0) = p e^{i(\omega t - kx)} where \omega^2=gk\tanh{kh}, in addition to the fact that the shear stress must vanish at the liquid interface \sigma_{xz}(z=0) = 0.
Since the horizontal wavenumber and frequency have been provided for us by the surface wave, it is natural to take the ansatz \phi = \Phi(z) e^{i(\omega t -kx)} and \psi = \Psi(z) e^{i(\omega t -kx)}, yielding:
\Phi'' + \left(\frac{\omega^2}{\alpha^2} - k^2\right) \Phi = 0 and \Psi'' + \left(\frac{\omega^2}{\beta^2} - k^2\right) \Psi = 0
Letting \nu_\alpha^2 = k^2 - \frac{\omega^2}{\alpha^2} and \nu_\beta^2 = k^2 - \frac{\omega^2}{\beta^2} (essentially the vertical wavenumber), the solution is simply
\Phi = a_1 e^{\nu_\alpha z} + a_2 e^{-\nu_\alpha z} (and similarly for \Psi). Because both \nu's are real (\nu_\alpha = \sqrt{k^2-\omega^2/\alpha^2} = k\sqrt{1-c^2/\alpha^2}, where c is the surface wave phase velocity, which is maybe 100x slower than the P or S wave velocities). We know that the correct solution doesn’t increase exponentially into the half-space, so we can assume a_2=0.
We therefore have some linear combination of the P and S waves, which propagates horizontally at the surface wave velocity and decays exponentially into the earth:
\phi = a e^{\nu_\alpha z} e^{i(\omega t -kx)} and \psi = b e^{\nu_\beta z} e^{i(\omega t -kx)}
To evaluate the boundary condition and determine a,b, we can write out the components of stress in terms of the potential functions (letting k_\alpha = \omega/\alpha and similarly with \beta for brevity):
\sigma_{zz} = \mu \left[(2k^2-k_\beta^2) \phi - 2i k\nu_\beta \psi \right]
\sigma_{xz} = \mu \left[-2ik\nu_\alpha \phi + (k_\beta^2-2k^2)\psi \right]
Factoring out the oscillatory term and evaluating this at z=0, then solving the resulting system of equations, one obtains
a = \frac{2k^2-k_\beta^2}{\mu R(k)} p and b = \frac{2ik\nu_\alpha}{\mu R(k)} p where R(k) = (2k^2 -k_\beta^2)^2 - 4k^2\nu_\alpha\nu_\beta is Rayleigh’s function and p is the wave pressure at the seafloor.
We can write the horizontal displacement in terms of the potentials:
u_x = \partial_x\phi - \partial_z\psi = -ik\phi - \nu_\beta\psi
Evaluating this at the seafloor (z=0, so e^{\nu_\alpha z} = 1):
u_x = \left[ \frac{k_\beta^2-2k^2 - 2 \nu_\alpha\nu_\beta}{\mu R(k)} \right] ikp e^{i(\omega t - kx)}
\varepsilon_{xx} = \partial_x u_x = -ik u_x = \left[ \frac{ k_\beta^2-2k^2 - 2\nu_\alpha\nu_\beta }{\mu R(k)} \right] k^2 p e^{i(\omega t - kx)}
Recalling the pressure boundary condition \sigma_{zz}(z=0) = p e^{i(\omega t - kx)}, you will note that the horizontal strain is in phase with the pressure. The pressure gradient associated with this surface wave is -ikp e^{i(\omega t - kx)}, which is 90 degrees out of phase from the strain. (Also notably, the opposite is true if you go evaluate the vertical displacement and strain – the vertical displacement is in phase with the pressure, which may be the source of confusion).
I will have to write the second and third parts later!