Suppose you've got two dice; what's the probability the sum of their rolls will add up to 4? You simply look at all the ways the values of the dice could sum to 4 (e.g. 1 and 3, 2 and 2 or 1 and 4), and add up their probabilities (in this case each is 1/36, so totalling 3/36 or 1/12). You could use a probability square to visualise this calculation.

This is the general procedure to sum two independent random variables. You've got two random variables X and Y; in the example above the dice. They each have some probability distribution function $$f_X$$ and $$f_Y$$ which encode how probable any particular outcome is. For the dice the probability distribution function is just $$f_X(x) = \frac{1}{6},\, x \in \{1,2,3,4,5,6\}$$ (that is the probability of any number between 1 and 6 is one-sixth). Then the probability distribution of their sum is then $$f_{X+Y}(x) = \sum_{y+z=x} f_X(y) f_Y(z)$$; that is for each pair of outcomes that sum up to x we multiply the probability of them occurring. In our dice example we calculated $$f_{X+Y}(4) = \frac{1}{12}$$; we could repeat the calcuation for all numbers between 2 and 12 to get the full probability distribution function.

The sum above can be rewritten in a slightly different form using a change of variables; $$f_{X+Y}(x) = \sum_{y} f_X(y) f_Y (x - y) =: (f_X * f_Y)(x)$$. This last expression is just noting this is the definition of a convolution. A shorthand way of writing this is $$f_{X+Y} = f_X * f_Y$$ (note that this is convolution and not multiplication!).

The convolution theorem says that the Fourier Transform maps convolutions into products. So in particular we can rewrite $$f_{X+Y} = \mathcal{F}^{-1} (\mathcal{F}(f_X) \mathcal{F}(f_Y))$$ This can be computationally convenient; a fast fourier transform can be more efficient than performing the sum in the convolution manually. To see this in action we can encode our dice example in R; we represent our probability density function as a vector which are the values of the function at 0, 1, 2, ...

# Vectors are 0 indexed, so probability of 0 as 0
# Probability at 1-6 as 1/6
# We need extra space at the end for higher values so we pad the end with 0s
x <- c(0, seq(1/6, 6), seq(0, 10))
y <- x

# Convoluiton using the fourier transforms and their inverse
z <- fft(fft(x) * fft(y), inverse=True)
# Remove the imaginary part (which should be 0)
z <- abs(z)
# Renormalise the probability to sum to 1
# (The fft introduces a constant multiplicative factor)
z <- z / sum(z)

# The result is what we would get from the convolution
# For example the value at 4 (the 5th element in the vector) is 3
print(round(z*36, 0.01))
#  0 0 1 2 3 4 5 6 5 4 3 2 1 0 0 0 0

These relations are also useful for theoretical calculations. For example the PDF of a standard normal distribution, $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ is a fixpoint of the Fourier transform. This with the scaling and shift relations make it easy to calculate the sum of normally distributed variables. In particular if $$X \sim \mathcal{N}(\mu_1, \sigma_1^2)$$ and $$Y \sim \mathcal{N}(\mu_2, \sigma_2^2)$$ then $$X + Y \sim \mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$$.