Assume $(p,q) = 1$. How the Chinese Remainder Theorem allows to "factorize" the exponential $$\sum_{u \mod pq} e\left( \frac{au}{q} + \frac{mu}{pq} \right)$$ into $$ \sum_{v \mod p} \sum_{w \mod q} e\left( \frac{m\bar{q}}{p} v \right) e\left( \frac{a + m\bar{p}}{q}w \right)$$ where $$e(x) = \exp(2i\pi x)?$$
There is a kind of crossed multiplicative/additive setting that I do not really understand here.
Given $v,w,p,q\in \mathbb Z$ with $(p,q)=1$, then \[ x\equiv v\mod (p)\] \[ x\equiv w\mod (q)\] has a unique solution modulo $pq$ by the Chinese Remainder Theorem. This solution is given mod $pq$ explcitly by \[ x\equiv v\overline qq+w\overline pp\mod (pq)\] where $\overline q$ is the inverse of $q$ mod $(p)$ and where $\overline p$ is the inverse of $p$ mod $(q)$. (To see this just explicitly check the value of it mod $p$ and mod $q$. Also note this really is well-defined mod $pq$ even if these inverses are defined only mod $p$ and $q$.) In other words, as $v$ and $w$ ranges over $p$ and $q$ then $x$ ranges over $pq$.
Therefore \[ \sum _{v\mod (p)}\hspace {5mm}\sum _{w\mod (q)}e_p(v)e_q(w)=\sum _{v\mod (p)}\hspace {5mm}\sum _{w\mod (q)}e_pq(v\overline qq+w\overline pp)\sum _{x=1}^{pq}e_pq(x)\]