This type of result, and the tools needed to prove it, would usually be discussed in a graduate level Analytic Number Theory text. My personal preference is Montgomery and Vaughn's "Multiplicative Number Theory I. Classical Theory."
Here is another way to see why this result is true. Lets examine the sum $\sum_{\begin{array}{c} n\leq x\\ n\equiv a\ (q) \end{array}}\omega(n).$ Since $\omega(n)=\sum_{p|n}1,$ the above becomes $\sum_{\begin{array}{c} n\leq x\\ n\equiv a\ (q) \end{array}}\sum_{p|n}1=\sum_{p\leq x}\sum_{\begin{array}{c} n\leq x,\ p|n\\ n\equiv a\ (q) \end{array}}1.$ This equals
$\sum_{\begin{array}{c} p\leq x\\ p\nmid q \end{array}}\sum_{\begin{array}{c} n\leq\frac{x}{p}\\ n\equiv p^{-1}a\ (q) \end{array}}1=\sum_{\begin{array}{c} p\leq x\\ p\nmid q \end{array}}\left(\frac{x}{pq}+O(1)\right) $
$=\frac{x}{q}\log\log x+\frac{B_{1}x}{q}+O\left(\frac{x}{\log x}\right),$ where $B_1$ is Mertens constant. From here, we can recover your asymptotic since $\log \log \left(\frac{x}{q}\right)\sim \log \log x$, and so $\sum_{n\leq x }\omega(nq+a)\sim\sum_{n\leq x }\omega(n)\sim x \log \log x.$