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
$$f(y\rightarrow0)\sim 2\sqrt{\frac{\pi}{y}} \exp\left\{-\frac{\pi^2}{4y}\right\}.$$
I am not sure how the Mumford claim is wrong, but simple
calculations demonstrate that it is. For example, PARI/GP code:
e(x) = exp(Pi*I*x);
Thn(n,z,t,a,b) = e((n+a)^2*t) * e(2*(n+a)*(z+b));
{Th(z,t,a,b) = suminf(n=0,Thn(n,z,t,a,b)) +
suminf(n=1,Thn(-n,z,t,a,b))};
a=.2; b=.2 ; t=1.*I;
print([Th((+a+.5)*t+b+.5,t,a,b), Th((-a+.5)*t-b+.5,t,a,b)])
[-0.766570202+4.83993375*I, 0.E-9+0.E-9*I]
It could be simple sign error which is easy to make.
Best Answer
The infinite product representation for $\vartheta_{1}(z, q)$ given by $$\vartheta_{1}(z, q) = 2q^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n}\cos 2z + q^{4n})\tag{1}$$ I hope you are familiar with the above result. Taking logs we get \begin{align}\log\vartheta_{1}(z, q) &= \log(2q^{1/4}) + \log\sin z + \log\prod_{n = 1}^{\infty}(1 - q^{2n})\notag\\ &\,\,\,\,\,\,\,\, + \sum_{n = 1}^{\infty}\log(1 - 2q^{2n}\cos 2z + q^{4n})\tag{2} \end{align} and differentiating with respect to $z$ we get \begin{align} \frac{\vartheta'_{1}(z, q)}{\vartheta_{1}(z, q)} &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}\sin 2z}{1 - 2q^{2n}\cos 2z + q^{4n}}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}\frac{q^{2n}(e^{2iz} - e^{-2iz})}{(1 - q^{2n}e^{2iz})(1 - q^{2n}e^{-2iz})}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}q^{2n}\left(\frac{e^{2iz}}{1 - q^{2n}e^{2iz}} - \frac{e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\frac{q^{2n}e^{2iz}}{1 - q^{2n}e^{2iz}} - \sum_{n = 1}^{\infty}\frac{q^{2n}e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{2iz}\}^{m} - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{-2iz}\}^{m}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{2imz}q^{2m(n - 1)}\right. \notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{-2imz}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}q^{2m}e^{2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right.\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{m = 1}^{\infty}q^{2m}e^{-2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}\frac{q^{2m}e^{2imz}}{1 - q^{2m}} - \sum_{m = 1}^{\infty}\frac{q^{2m}e^{-2imz}}{1 - q^{2m}}\right)\notag\\ &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}}{1 - q^{2n}}\sin 2nz\tag{3}\end{align}
The above proof is taken from one of my blog posts (see equation $(25)$ of that post).