One approach is to use the Polya-Vinogradov method: i.e. using a Fourier transform mod $2q$, find $c_m$ such that $$\sum_{m=1}^{2q} c_m e \left( \frac{mn}{2q} \right)= \begin{cases} 1 & \textrm{if } 1 \leq n \leq q \\ 0 & \textrm{if } q+1 \leq n \leq 2q \end{cases}$$ so that $$t(\chi) = \sum_{n=1}^{2q}  \chi(n) e\left( \frac{n}{2q} \right)\sum_{m=1}^{2q}c_m e \left( \frac{mn}{2q} \right) = \sum_{m=1}^{2q} c_m \sum_{n=1}^{2q} \chi(n) e\left( \frac{n (m+1)}{q} \right) $$

and the inner sum vanishes if $m+1$ is odd and otherwise is a standard Gauss sum mod $q$ and can be bounded in the usual way. This will give a bound of the rough form $O(\sqrt{q} \log q)$.