Ok, you can, after some calculations see that the absolute value of your sum is less than or equal to $\sum_{n=-\infty}^\infty e^{\pi(-tn^2 + 2n|z|)}$ where $t = im(\tau) >0.$
The part of the sum where $n<0$ will be less than some finite constant depending only on $\tau,$ so we only need to worry about $n>0.$ (This you need to prove.)
Ok, so, divide that sum into two parts:
Let $N$ be the least integer greater than $4|z|/t$ (where t = im(\tau) >0.)
If $n\geq N$ we have $nt/4 > |z|.$ This will imply that $-tn^2 + 2n|z| \leq -n^2t/2.$
Use this on the second sum, $N...\infty$ and use some other estimations on the first part.
Edit: Continuation:
First, you should see that $\sum_{n=-\infty}^0 e^{\pi(-tn^2 + 2n|z|)} \leq \sum_{n=-\infty}^0 e^{-\pi t n^2} $ and this clearly converges to something that only depends on $\tau.$ (The exponent will diverge to $-\infty$ rapidly, so the sum must converge.)
Now, if $n\geq N$ then
$\sum_{n=N}^\infty e^{\pi(-tn^2 + 2n|z|)}\leq \sum_{n=N}^\infty e^{\pi(-n^2 t/2)} $ which converges to a finite number independent of $z$.
Now, you only need to do something about the last part: $\sum_{n=0}^{N-1} e^{\pi(-tn^2 + 2n|z|)}$
(Hint: use $N-1< 4|z|/t$ which follows from the conditions above).
Edit: Continuation:
Ok, choose $r \in \mathbb{R}$ so that $\theta(r,\tau) \neq 0.$ By quasi-periodicity, we have for integers $m$ that $\theta(r+m\tau,\tau) = e^{-i\pi m^2\tau - 2im \pi r}\theta(r,\tau)$ so by taking absolute values on each side, we see that $|\theta(r+m\tau,\tau)| = e^{t\pi m^2}|\theta(r,\tau)|$ where $t = im(\tau)>0.$ Hence, $\lim_{m \rightarrow \infty} \frac{|\theta(r+m\tau,\tau)|}{e^{t\pi m^2}} = |\theta(r,\tau)| > 0.$ Thus, the order is at least 2.