Fourier convolution of a histogram Notation: 


$\mathcal F\left\{a\right\}$ denotes applying the discrete forward Fourier transform to the histogram $a$. Similarly for $\mathcal F^{-1}\left\{a\right\}$ and the discrete inverse Fourier transform.
Capitals represent the discrete fast Fourier transform of the lower-case counterpart, e.g. $A = \mathcal F\left\{a\right\}$. 
The convolution of two histograms $f$ and $g$ is denoted $f\star g$.


Say I have histograms $a$ and $b$ with $a=[1,2,3]$ and $b=[4,5,6]$ (the heights) and the edges given by $e=[0,1,2,3]$. I.e. For the histogram $a$, there is one entry in the interval $[0,1)$, two entries in the interval $[1,2)$, and three entries in the interval $[2,3]$.

In order to perform a linear convolution as opposed to circular, $a$ must be padded such that len($a$) = len($a$) $+$ len($b$) $-1$. Similarly for $b$. 

The convolution theorem states: 
$$\mathcal F\left\{a\star b\right\} = \mathcal F\left\{a\right\} \cdot \mathcal F\left\{b\right\}$$ 

So if $c = a\star b$ and $d = \mathcal F^{-1}\left\{A\cdot B\right\}$, then $c = d$.

On performing the Fourier convolution of the $a$ and $b$ mentioned above,  I obtain a result if $d = [4,13,28,27,18]$.

My questions: 


Where does $c$ sit in terms of bins? 
Do I need to expand the bin edges, $e$, in some way? 
In which direction should I expand them? 
Should the extra bins have the same width as the originals (in this case, $1$)? 


Intuition tells me that there should be an extra bin on either side of $e$, of the same width. 

Furthermore, if I was to crop $c$ such that it has the same length as the unpadded kernel or the signal, where would I make the cut? Is this even possible?
Actually, it does not really matter, if a or b are histograms or time domain sequences. However, I suppose you want to convolve to histograms to get the probability distribution of a random variable that is the sum of two variables. Then, its PDF is the convolution of the PDF of the original variables. But, considering the convolution of two time-domain sequences might make it easier to understand. 
The convolution theorem states fft(ab)=fft(a)*fft(b), where * denotes the N-point circular convolution and a and b are of length N points. Your statement of the convolution theorem fft(ab) = fft(a) . fft(b) is not correct.
As I said, the convolution theorem in discrete domain corresponds to circular convolution. Most probably, you want to have a linear convolution. Then, you need to zero-pad a and b before applying the convolution theorem. 
If the original bins range from 0-3 (left-side of the interval), then the bin edges after convolution would range from 0-6, because the (linear) convolution has a length of (N+N-1) samples, where N is the length of the sequence (which is 4 in your case). You can also understand this from the fact that the convolution of the histogram is the PDF of the sum of both random variables: The minimum sum would be 0, the maximum sum would be 3+3=6.


For more information on circular convolution, you might have a look at an article I wrote or the Wikipedia entry.