I'll show how the dispersion relation for surface water waves can be derived.
The words about "resonance" in the body of the question are true in the broad sense that any wave can be regarded as a kind of resonance, loosely speaking. The question in the title is more specific, so I'll focus on that. A wave is called dispersive if its speed depends on its wavelength $\lambda$. In terms of the angular frequency $\omega=2\pi f$ and wavenumber $k=2\pi/\lambda$, this condition means that $\omega$ is not proportional to $k$. The goal is to understand why surface water waves have this property.
Inputs
Let $\mathbf{x}=(x,y,z)$ be a point in space, either on or below the surface of the water, and let $\mathbf{v}(\mathbf{x})$ be the velocity of the water at that point. Take $z$ to be the vertical direction, and let $h(x,y,t)$ be the height of the water's surface relative to what it would be with no waves. The wave is governed by two forces:
Gravity, which tries to keep the surface low.
Surface tension, which tries to keep the surface flat.
According to refs 1 and 2, the force of gravity is the dominant one for long wavelengths $\lambda\gg 2\text{ cm}$ (called gravity waves), and the surface tension dominates for short wavelengths $\lambda\ll 2\text{ cm}$ (called capillary waves). Both limits exhibit dispersion. I'll keep the wavelength $\lambda$ arbitrary, but I'll use these approximations:
The water is incompressible: $\nabla\cdot\mathbf{v}=0$. The density $\rho$ is constant and uniform.
The water is irrotational: $\nabla\times\mathbf{v}=0$.
The bottom of the body of water is flat (no underwater mountains).
The changes in height $h$ (the amplitude of the wave) are small enough to justify using the linear approximation. This approximation allows us to ignore the difference between the time-derivative of $\mathbf{v}$ at a fixed point in space and the "material" time-derivative of $\mathbf{v}$ at a point that moves along with the flow.
Analysis
The irrotational condition implies that we can write $\mathbf{v}$ as the gradient of a scalar function $\phi$: $$ \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfx}{\mathbf{x}} \bfv=\nabla\phi. $$ Then the incompressibility condition implies $$ \nabla^2\phi=0. \tag{I} $$ At the surface, the $z$-component of $\bfv$ must match the time-derivative of the height-function $h$: $$ \nabla_z\phi=\frac{dh}{dt} \hspace{1cm} \text{at }z=h. \tag{H} $$ At the bottom of the body of water, the $z$-component of $\bfv$ must be zero: $$ \nabla_z\phi=0 \hspace{1cm} \text{at }z=-D, \tag{D} $$ where $D$ stands for "depth." At any point under the surface, the water's acceleration is determined by the forces of gravity and pressure: $$ \frac{d\bfv}{dt}+g\nabla z+\frac{\nabla P}{\rho}=0, $$ where $g$ is the acceleration of gravity, $P$ is the pressure, and $\rho$ is the density (which we're assuming is constant and uniform — the incompressibility condition). Note that $\nabla z$ is just a convenient way of writing the vertical unit vector. Write this equation in terms of $\phi$ to get $$ \nabla\left(\frac{d\phi}{dt}+g z+\frac{P}{\rho}\right)=0, $$ which implies $$ \frac{d\phi}{dt}+g z+\frac{P-P_0}{\rho}=0 $$ where $P_0$ is an integration constant that we can interpret as the pressure at the surface. The jump in pressure at the surface is determined by the surface tension: $$ P-P_0 = -T \nabla^2 h $$ where $\nabla^2 h$ measures the curvature of the surface and $T$ is a constant relating the tension to the degree of curvature. (Surface tension tries to make the surface flat, so more curvature means more tension.) Use that in the preceding equation to get $$ \frac{d\phi}{dt}+g h-\frac{T}{\rho}\nabla^2 h=0 \hspace{1cm} \text{at }z=h. $$ Recall that $h$ is independent of $z$, so this can also be written $$ \frac{d\phi}{dt}+g h-\frac{T}{\rho}(\nabla_x^2+\nabla_y^2) h=0 \hspace{1cm} \text{at }z=h. \tag{1} $$ To express this entirely in terms of $\phi$, take the time-derivative of (1) and use (H) to get $$ \frac{d^2\phi}{dt^2}+g \nabla_z \phi -\frac{T}{\rho}(\nabla_x^2+\nabla_y^2) \nabla_z \phi =0 \hspace{1cm} \text{at }z=h. \tag{2} $$ Now consider a wave moving in the $x$-direction: $$ \phi(\bfx)=f(z)\sin(\omega t-kx), \tag{A} $$ where $f$ is a function to be determined. Use this ansatz in (2) to get $$ -\omega^2 f+g f'+ \frac{T}{\rho}k^2 f' = 0 \hspace{1cm} \text{at }z=h, \tag{3} $$ where $f'(z)=df/dz$. Use the ansatz (A) in the incompressibility condition (I) to get $$ f''-k^2f =0, $$ and combine this with the bottom boundary condition (D) to infer $$ f(z)\propto \cosh(kz+kD). $$ Use this in (3) to get the final result $$ \omega^2 \approx \left(gk+\frac{T}{\rho}k^3\right)\tanh(kD), \tag{R} $$ where the small-amplitude approximation was used again to simplify the argument of the tanh function. This is the dispersion relation for surface water waves, under the approximations listed above.
Equation (R) shows that $\omega$ is generally not proportional to $k$, so surface water waves are generally dispersive: their speed depends on their wavelength. The speed (phase velocity) is $\omega/k$, so equation (R) shows that in the deep-water approximation ($\tanh(kD)\approx 1$), longer wavelengths travel faster when gravity is the dominant restoring force (gravity waves), and shorter wavelengths travel faster when surface tension is the dominant restoring force (capillary waves).
References:
Kartashova (2009), "Nonlinear resonances of water waves," Discrete and Continuous Dynamical Systems Series B 12:607-621 (https://www3.risc.jku.at/publications/download/risc_3822/resonances_6.pdf)
Mei (2004), "Chapter four: waves in water," notes for an MIT course in wave propagation (http://web.mit.edu/1.138j/www/material/chap-4.pdf)