[proofplan]
We expand the finite [Fourier transform](/page/Fourier%20Transform) of the representation function using the definition of $r_{A+B}$. Since every pair $(x,y)\in G\times G$ contributes to exactly one residue class $z=x+y$, the sum over $z$ collapses to the phase attached to $x+y$. The remaining double sum separates into a product because the additive character satisfies $e(-(a(x+y))/q)=e(-ax/q)e(-ay/q)$.
[/proofplan]
[step:Rewrite the representation function as a finite convolution]
Let $G:=\mathbb Z/q\mathbb Z$. The indicator functions are the maps
\begin{align*}
\mathbb 1_A:G\to\{0,1\},\qquad x\mapsto \begin{cases}1,&x\in A,
\end{align*}
\begin{align*}
0,&x\notin A,\end{cases}
\end{align*}
and
\begin{align*}
\mathbb 1_B:G\to\{0,1\},\qquad y\mapsto \begin{cases}1,&y\in B,
\end{align*}
\begin{align*}
0,&y\notin B.\end{cases}
\end{align*}
For each $z\in G$, the definition of $r_{A+B}$ gives
\begin{align*}
r_{A+B}(z)=\sum_{\substack{x\bmod q, y\bmod q, x+y=z}}\mathbb 1_A(x)\mathbb 1_B(y).
\end{align*}
Equivalently, this is the finite convolution identity
\begin{align*}
r_{A+B}(z)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)\mathbb 1_{\{z\}}(x+y),
\end{align*}
where $\mathbb 1_{\{z\}}:G\to\{0,1\}$ is the indicator function of the singleton $\{z\}$.
[/step]
[step:Expand the Fourier transform and collapse the sum over the output residue]
Fix $a\in G$. By the Fourier transform convention in the statement,
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{z\bmod q}r_{A+B}(z)e\left(-\frac{az}{q}\right).
\end{align*}
Substituting the convolution expression for $r_{A+B}(z)$ and using finiteness of $G$ to interchange the sums gives
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)\sum_{z\bmod q}\mathbb 1_{\{z\}}(x+y)e\left(-\frac{az}{q}\right).
\end{align*}
For fixed $x,y\in G$, exactly one residue class $z\in G$ satisfies $z=x+y$, so the inner sum is
\begin{align*}
\sum_{z\bmod q}\mathbb 1_{\{z\}}(x+y)e\left(-\frac{az}{q}\right)=e\left(-\frac{a(x+y)}{q}\right).
\end{align*}
Therefore
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)e\left(-\frac{a(x+y)}{q}\right).
\end{align*}
[guided]
Fix a residue class $a\in G$. The Fourier transform of $r_{A+B}:G\to\mathbb N\cup\{0\}$ is, by definition,
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{z\bmod q}r_{A+B}(z)e\left(-\frac{az}{q}\right).
\end{align*}
The purpose of introducing $r_{A+B}$ is that it counts pairs. For a fixed residue class $z\in G$, the value $r_{A+B}(z)$ is the number of pairs $(x,y)\in A\times B$ with $x+y=z$. Writing this count with indicator functions gives
\begin{align*}
r_{A+B}(z)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)\mathbb 1_{\{z\}}(x+y).
\end{align*}
Substituting this identity into the Fourier transform gives
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{z\bmod q}\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)\mathbb 1_{\{z\}}(x+y)e\left(-\frac{az}{q}\right).
\end{align*}
All sums are finite because $G$ has exactly $q$ elements, so we may rearrange them without any convergence issue:
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)\sum_{z\bmod q}\mathbb 1_{\{z\}}(x+y)e\left(-\frac{az}{q}\right).
\end{align*}
Now fix $x,y\in G$. In the group $G$, there is exactly one element equal to $x+y$. Hence the singleton indicator $\mathbb 1_{\{z\}}(x+y)$ is equal to $1$ for precisely the residue class $z=x+y$ and is equal to $0$ for all other residue classes. Therefore the inner sum over $z$ collapses to the single surviving term
\begin{align*}
\sum_{z\bmod q}\mathbb 1_{\{z\}}(x+y)e\left(-\frac{az}{q}\right)=e\left(-\frac{a(x+y)}{q}\right).
\end{align*}
Thus
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)e\left(-\frac{a(x+y)}{q}\right).
\end{align*}
[/guided]
[/step]
[step:Factor the double sum into the two Fourier transforms]
The exponential character is multiplicative with respect to addition in the argument, so for every $x,y\in G$,
\begin{align*}
e\left(-\frac{a(x+y)}{q}\right)=e\left(-\frac{ax}{q}\right)e\left(-\frac{ay}{q}\right).
\end{align*}
Substituting this into the previous formula gives
\begin{align*}
\widehat{r_{A+B}}(a)=\sum_{x\bmod q}\sum_{y\bmod q}\mathbb 1_A(x)\mathbb 1_B(y)e\left(-\frac{ax}{q}\right)e\left(-\frac{ay}{q}\right).
\end{align*}
Since the first factor depends only on $x$ and the second factor depends only on $y$, the finite double sum factors:
\begin{align*}
\widehat{r_{A+B}}(a)=\left(\sum_{x\bmod q}\mathbb 1_A(x)e\left(-\frac{ax}{q}\right)\right)\left(\sum_{y\bmod q}\mathbb 1_B(y)e\left(-\frac{ay}{q}\right)\right).
\end{align*}
By the Fourier transform convention, the two factors are $\widehat{\mathbb 1_A}(a)$ and $\widehat{\mathbb 1_B}(a)$. Hence
\begin{align*}
\widehat{r_{A+B}}(a)=\widehat{\mathbb 1_A}(a)\widehat{\mathbb 1_B}(a).
\end{align*}
Since $a\in G$ was arbitrary, the identity holds for every $a\in G$.
[/step]
