As @TerryTao stated correctly in the comments, a bound on on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially:

First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$:

$$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponenent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$.

To derive the above formula, Jenssen refers to an exercise in the book Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library). He essentially solves this exercise which leads to the above expression.

The above formula should agree with the one derived in the answer of @მამუკა ჯიბლაძე.

As @TerryTao stated correctly in the comments, a bound on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially:

First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$:

$$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$.

To derive the above formula, Jenssen refers to an exercise in the book Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library). He essentially solves this exercise which leads to the above expression.

The above formula should agree with the one derived in the answer of @მამუკა ჯიბლაძე.

As @TerryTao stated correctly in the comments, a bound on on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially:

First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$:

$$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponenent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$.

To derive the above formula, Janssen refers to an exercise in the book Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library). He essentially solves this exercise which leads to the above expression.

The above formula should agree with the one derived in the answer of @მამუკა ჯიბლაძე.

As @TerryTao stated correctly in the comments, a bound on on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially:

First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$:

$$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponenent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$.

To derive the above formula, Jenssen refers to an exercise in the book Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library). He essentially solves this exercise which leads to the above expression.

The above formula should agree with the one derived in the answer of @მამუკა ჯიბლაძე.