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
$$\sum_{n=N}^\infty e^{\pi(-tn^2 + 2n|z|)}\leq \sum_{n=N}^\infty e^{\pi(-n^2 t/2)}
\leq \sum_{n=0}^\infty e^{\pi(-n^2 t/2)}
$$
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.
UPD: the previous version contained a square which shouldn't be there.
Actually, your function is even more simply expressed in terms of $\vartheta_4$-function. Also, I prefer this notation in which
$$f(y)=\vartheta_4(0,e^{-y})=\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr).$$
I.e. I use the convention $\vartheta_k(z,q)=\vartheta_k(z|\tau)$.
Then, to obtain the asymptotics as $y\rightarrow 0^+$, we need two things:
Jacobi's imaginary transformation, after which the transformed nome and half-period behave as $q'\rightarrow0$, $\tau'\rightarrow i\infty$ (instead of $q\rightarrow1$, $\tau\rightarrow0$):
$$\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr)=\sqrt{\frac{\pi}{y}}\vartheta_2\Bigl(0\Bigr|\Bigl.\frac{i\pi}{y}\Bigr).$$
Series representations for theta functions (e.g. the formula (8) by the first link), which implies that
$$\vartheta_2(0,q')\sim 2(q')^{\frac14}$$
as $q'\rightarrow 0$. Note that you can also obtain an arbitrary number of terms in the asymptotic expansion if you want.
Taking into account the two things above, we obtain that the leading asymptotic term is given by
$$f(y\rightarrow0)\sim 2\sqrt{\frac{\pi}{y}} \exp\left\{-\frac{\pi^2}{4y}\right\}.$$
Best Answer
The statement "Theta Functions are elliptic analogues of the exponential function" is not correct because it does not specify in which sense they are elliptic analogues. It is more accurate to state that the Jacobi elliptic functions (which are quotients of theta functions) are the elliptic analogues of trigonometric functions in the following sense. The Wikipedia article Jacobi elliptic functions states
The Jacobi elliptic functions are doubly periodic which generalizes the single periodicity of the trigonometric functions which they reduce to. In fact, the current notation, along with the notation used by Jacobi is as follows: $$\textrm{sn}(u, k) := \sin( \textrm{am}(u, k)). $$ $$\textrm{cn}(u, k) := \cos( \textrm{am}(u, k)). $$ $$\textrm{dn}(u, k) := \Delta( \textrm{am}(u, k)). $$ When $\, k=0, \,$ then $\, \textrm{am}(u, 0) = u \,$ and this leads to $$ \textrm{sn}(u, 0) = \sin(u), \, \textrm{cn}(u, 0) = \cos(u), \, \textrm{dn}(u,0) = 1. $$ When $\, k=1, \,$ then $\,\textrm{am}(u,1) = \textrm{gd}(u),\,$ and this leads to $$ \textrm{sn}(u, 1) = \tanh(u), \, \textrm{cn}(u, 1) = \textrm{dn}(u, 1) = \textrm{sech}(u). $$ Thus, the Jacobi elliptic functions are a common generalization of the circular and hyperbolic trigonometric functions.