Why can you hear sound over a wall?

It is not quite clear to me what you are asking, but I would refer to simulations like on PhET: Wave Interference.

There you can switch between sound waves, surface waves on water (ripple tank) or light waves.

There are two important things to realize:

• The wavelength of sound waves is of the order of 1 meter, comparable to the height of walls.
• We experience sound level as a logarithmic measure of the intensity, and for example even a tenfold reduction in power reduces the sound level by only 10 dB, for example from 60 dB to 50 dB.

If the propagation of sound waves and light waves are governed by exactly the same wave equation: $$frac{partial^2 f}{partial t^2}=c^2 nabla^2f,$$ where $c$ is the propagation speed of the waves and $f$ is the thing that propagates (say, pressure for a sound wave or electric field for a light wave) why can you hear but not see what’s happening on the other side of a wall?

There are two main reasons, the more important of which Pieter already pointed out: the wavelength of sound waves is a million times bigger. Sound in the human auditory range diffracts around a 10-m wall the same way light we can see diffracts around a “wall” with a height of 0.01 mm. (The speed is also different by a factor of about a million, but that doesn't change the spatial characteristics of how the waves behave as they travel from source to you, only how long that trip takes.)

The other, subtler reason is that we process auditory and visual signals very differently.

All that really matters to your brain about the pressure signals coming into your ears is the spectrum—the amount of energy at each frequency. So the signal can be diffracted (or bounce off irregular surfaces or whatever) without affecting your ability to extract information from it. (Incidentally, in many cases scattering is more important than diffraction—when you hear a voice from the other room, much of that sound energy reaches you by bouncing off a wall or two.)

Your eyes, on the other hand, don’t care about the spectrum—color is just some sort of average of all the constituent frequencies—but they do need light rays originating from various positions on some object you'd like to see to be traveling predictably along the lines from those positions. Otherwise no image of the object is formed. Diffraction (or bounces off rough surfaces) messes this up; even if the light still enters your eye, the image that would have been formed on your retina by rays traveling direct from the source is now scrambled, making it impossible for your brain to extract from the light the information it would have gotten from that image.

If you can follow this logic you should get a feel for what happens...

Waves that aren't physically confined have a tendency to spread.

Imagine using a stick to agitate water in a tank- you will make waves that travel outwards in a circle, spreading over an increasing area.

If the tank is a long channel, maybe a foot across and say twenty feet long, when the waves try to spread they are constrained by the walls, so their movement is directed along the channel.

Now imagine the channel opens out into a pond at the far end. When the waves arrive there they are no longer constrained by the walls so they open out again.

So returning to your question, when something makes a sound on the other side of a wall, the sound waves try to spread in all directions. The sound that makes it over the wall is like the water wave leaving the channel- it spreads out beyond the wall.

When you consider the effect in more detail you can show (I won't try to do it here) that the extent of the spread when the wave moves out of the channel into the pond depends upon the width of the opening at the end of the channel compared with the wavelength of the wave. When the two are about the same the spread is greatest. That explains why typical audible sounds (which have wavelengths of perhaps a few metres down to a few centimetres) spread around everyday gaps and obstacles.

An intuitive picture can be obtained using Huygen's construction. Although it is not a quantitative method, it is very useful for intuition. It says that every point on a wavefront acts as a trigger for a secondary spherical wavefront emanating from it. The superposition of these secondary wavefronts results in the whole wavefront moving forward - and so on. With this picture of wave propagation, imagine what happens if a section of wavefront is screened out by a wall -- the secondary spherical wavefronts right on the edge simply radiate into the geometric shadow region. That is diffraction. It is simply a consequence of the tendency of waves to spread.

In fact, a mathematically rigorous version of this argument is precisely the Kirchhoff approximation. It is still an approximation though. A truly rigorous geometric picture of diffraction, one that is derived directly from the wave equation without any additional postulates, does not exist. So based on our current understanding of wave physics, this is probably as far as you can go before getting the answer "solve the wave equation."