Hi Rüdiger
First things first.
The Fourier Transform is an operation that in its core what it does it trying to modelate your data through sine and cosine functions. The value of the FT, at a given point, is the amplitude of the sine or cosine that better fits the data (or, to be more precise, is the projection of your data to the space where sine and cosines live. To discriminate between sines and cosines the imaginary number is used. So, if you have, for example 1+0·i, then you are using just a cosine. On the other hand, a 0+1·i value tells you that this is only a sine. Mixed terms tell you that there is a mixture of a sine and a cosine, both of the same frequency. Usually the frequency of the sines and cosines are plotted from the center of the image, increasing theyr value. Also, since negative frequencies are the same as positive ones, all the data is mirroed; the forier transform is simetrical (in the diagonal). Going back to the mixture of a sine and a cosine for a given frequency, a complex number "a+b·i" may algo be written as: "c·e^(i·phi)", using Euler's theorem. phi is the phase, and c is the magnitude. The magnitide is like the length of the vector, the sum of the square components. The phase tells you the fraction of sines and cosines the signal has. It has the same meaning as change of phase in the senosoid signal. Let's take a cosine: f(x) = cos(f*x). If you add a phase: f(x) = cos(f*x + phi), the function will start at a different place. If the phase is 90 degrees (pi/2), then the cosine is equivalent to a sine without any phase. So, the phase information gives you an idea of the placement of the signal, although is not really straigthforward to interpret. I may give you an example of this if you want, for a very simple case. Anyway, for most applications, phase information is ignored, or just never touched. What we modify or inspect is the magnitude information.
Now, about the size of the image in the fourier space. The Fourier Transform is a very expensive operation. It is proportional to N^2 (N is the total number of pixels). To make it more manegeable, a very smart procedure was designed, the Fast Fourier Transform (FFT), which is only proportional to N·log(N). This is really a lot faster, but the problem with the first (and faster) FFT is that it needed the image to be a perfect multiple of 2^n. If you have another geometry, you cannot use the FFT. So, to overcome this, images are zero padded (zeros are added outside the boundaries of the image, to make is larger) until you reach the 2^n requisite. In modern implementations of the FFT, you don't have to go to 2^n, but any integer would work just fine too (for example 5^n), although not as fast. So, going back on track, the result of this is first: you have to operate with larger images, and second: these extra pixels, padded with zeros, create an artificial boundary on the images. Images are modeled by the standard FT as a periodical signal (that's why we try to use sines and cosines). By adding those zeroes, a very sharp edge is created in the image, at every boundary. To represent a sharp edge we need a lot (all) of sines and cosines, with different frequencies. That's why a cross is created in the fourier space. There are all the senosoids needed to represent those edges (both vertical and horozontal... we do not need diagonal sinosoids).
A final though about noise inspection in the fourier space. It is only benefical if you think that you have some periodical patterns in there. These patterns may be modelled by sinosoids, so you should end with a peak (and it's mirror) in the fourier space. If you have verticall lines, you'll get some dots in the horizontal line. If the pattern is not closelly fitted by sinosids, you'll end with spreaded peaks (you'll need more sinosoids to reconstruct the original pattern). If you don't have any periodical pattern, or clear peaks in the fourier space, then most probably a noise reduction scheme in the normal image space would work way much better.
Hope this helps.
