Why are there only a few high frequency components in the Discrete Cosine Transform?
There are two parts to the answer.  One has to do with the nature of sharp edges, the second has to do with the how imaging systems (lenses and the human eye) work.

As @PeterK's example demonstrates artificial (human created) images tend to have large regions of one color with sharp transitions between the regions.  This is similar to a square wave.  But if you look at the fourier transform of a square wave it will be a decreasing sum of the odd frequencies:
$$
\sum_{n=1, 3, 5, ...}\frac{1}{n} \sin{ n \omega t}
$$

Images of the natural world also often have regions of slowly changing color with sharp transitions where one object overlaps another.  There can actually be a lot of high frequency "texture" in an image (grass, leaves, a pebbly beach, waves on water that is relatively far away, whispy clouds) but the sharp edges of the occlusions (overlaps) obey the same square-wave fall-off in frequencies beyond the fundamental frequency (the average distance between tree trunks, blades of grass, etc.)

The second part of the answer is that imaging systems (camera lenses/eyes) can not make out features smaller than a certain minimum angular resolution.  Images outside of the imaging systems depth of field will be blurred because lenses have focal points.  But even objects inside the depth of field will be blurred.  This is because the aperture through which the light must pass must be reasonably large (several mm) in order to let in enough light.  But this means that point-sources of light actually create a circle of confusion that essentially acts as a low-pass filter.

Because of this the human eye, for example, is more sensitive at some frequencies than others.  In the below image the contrast differences across horizontal scans in the image are the same, but your eye is more sensitive to mid-frequency contrast differences than high-frequency contrast differences.  (You may have to adjust the distance from the image to your eye to see the effect line up with the black line drawn on the image.)