We all know that as waves approach the shallow shores, the waves begin to form a characteristic shape. The upper portion of these breaking waves appears to curl forward and downwards over the bottom segment of the wave, before breaking into "white wash". The image below illustrates what this characteristic shape looks like:
So why do waves form this characteristic breaking shape as they approach the shallow shores?
The physical process you describe is known as wave shoaling.
At the basic level, waves propagating into shallow water become shorter and higher, and consequently, steeper. In shallow water, the water particles near the crest move forward faster than those below them. Similarly, the particles near the trough move backward faster than those above them. This causes strong shearing of the near-surface body of water, eventually forming a plunging breaker, or a surf wave. For small-slope (linear) and inviscid (no friction) waves, the above is a consequence of the bottom boundary condition for the water velocity to be zero at the sea floor.
There are two fundamental and related properties of water waves that contribute to shoaling. One is the wave frequency $\omega$ remaining constant as the depth $d$ decreases. Think of this as the conservation of wave crests at any fixed point. However, the wavenumber $k$ (wavelength $\lambda$) must increase (decrease) with decreasing depth, as per the dispersion relationship of water waves (neglecting the effects of viscosity):
$$\omega^{2} = gk\tanh{(kd)}$$
where $g$ is gravitational acceleration. In shallow water, the dispersion relationship reduces to:
$$\omega = \sqrt{gd}k $$
and phase speed $C_p$ and group velocity $C_g$ are both proportional to the square root of the water depth:
$$ C_p = \dfrac{\omega}{k} = \sqrt{gd} $$
$$ C_g = \dfrac{\partial\omega}{\partial k} = \sqrt{gd} $$
Individual wave crests propagate with phase speed $C_p$. Wave groups and wave energy propagate with group velocity $C_g$. Thus, waves entering shallow water become shorter and slower. The second important property leading to shoaling is the conservation of wave energy flux, which is proportional to the group velocity and wave energy:
$$\dfrac{\partial(C_{g}E)}{\partial x}=0$$
Because the group velocity decreases, the wave energy (read: height) must increase (locally). This causes the waves to grow in height as they enter shallow water. As pointed out by @IsopycnalOscillation in a comment above, the separation of water particles from the wave crests happen because the individual orbital velocities at the top of the crest exceed by far the phase speed of the wave.
Although bottom friction is significant and non-negligible in shallow water, it does not cause shoaling. Mathematically, wave shoaling can occur for completely inviscid and small-slope (linear) waves that propagate into water of decreasing depth.
One thing that I would add is a discussion of a physical parameter that is a simple measure of whether a wave is going to topple or not. The Froude number is defined as the ratio of the maximum absolute fluid velocity, $U$, and the wavespeed, $c$:
$F = U/c$
Because of the mechanism explained by IRO-bot,
In shallow water, the water particles near the crest move forward faster than those below them. Similarly, the particles near the trough move backward faster than those above them.
the highest velocities are encountered at the top of the wave and therefore a good rule of thumb is, if the surface fluid velocity aproaches the wave speed, i.e $F \to 1$, then the wave will topple over.
The question is what values of $F$ give rise to that perfect tube feeling that surfers yearn for and so perfectly exemplified by your picture.
Of course, since $F > 1$ implies that the wave amplitude is large, linear theory is no longer helpful. We may have to wait until we can get some definite answers from DNS 3D simulations of oceanic scale wave breaking phenomena.
