A monochromatic beam of, say, red light appears red to the human eye. However, if we chop the beam with a mechanical shutter, we get a square wave. Fourier's theorem predicts that this beam will no longer be monochromatic as a square wave contains an infinite number of frequencies. So why does chopped red light, still appear red? What happens as the frequency of chopping increases?
This question addresses the principle by which supercontinuum lasers work. A short laser pulse consisting of just a few optical cycles does indeed contain a very broad distribution of Fourier modes with a wide range of frequencies, each of which may be considered to be an infinite plane wave. When the coherent pulse is created, however, these Fourier modes are in a specific phase relationship so that they cancel each other out before the pulse "begins" and after it "ends". During the pulse, only the frequency of the carrier wave is evident, and the pulse has that color.
The phase relationship between the Fourier modes can be changed by passing the light through a dispersive medium that delays the phase of the different modes by different amounts. Changing the phase relationship changes the temporal evolution of the superposition. The different modes no longer coincide in just the right way to make a temporal pulse of a particular color over a particular time interval. With enough dispersion, the monochromatic pulse can be transformed into a polychromatic quasi-continuous beams. This is how a supercontinuum laser works. In a typical implementation, a 100 fs pulse at 790 nm is transformed by passage through a dispersive optical fiber into a broad-band beam covering the wavelength range from 400 nm to 1400 nm. These beams appear white.
So, I suppose the short answer is that short pulses do have rich Fourier spectra and that the phase relationship among the coherent Fourier components can be modified to make these many colors apparent.
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