I was able to explain this behavior in the special case $\gamma = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$. I am guessing a similar argument could be made more generally but I don't have time to work out all the details.
I start by considering the auxiliary function $$h(x) = \int_1^{-1} (u+x)^{-\frac{3}{2}} \theta_3(u) du$$
where $\theta_3(z) = \sum_{n=-\infty}^\infty e^{i\pi n^2 z}$. This integral defines an analytic function on a Riemann surface consisting of two upper half-planes $U_1, U_{-1}$ glued to four lower half-planes $L_1, L_{-1}, L_1', L_{-1}'$, where $U_i$ is glued to $L_i$ along the real interval $(-\infty, -1)$, $U_i$ is glued to $L_{-i}$ along $(1, \infty)$, and $U_i$ is glued to $L_i'$ along $(-1, 1)$. Some easy properties of $h$ are that $h(x) = (-ix)^{-\frac{3}{2}} h\left(-\frac{1}{x}\right)$ and $h(x) = i\overline{h(-\overline{x})}$. The values of $h$ on the sheets labelled with a $1$ are the negatives of its values on the corresponding sheet labelled with a $-1$.
If we assume $\operatorname{Im}(x) < 0, \operatorname{Re}(x) < -1$, and take $\arg(u+x) \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, then fixing $b$ in the upper half-plane with $\operatorname{Re}(b) < -\operatorname{Re}(x)$, we have
$$h(x) = \left(\int_1^b + \int_b^{-1}\right) (u+x)^{-\frac{3}{2}} \theta_3(u) du$$
Writing $u = 1-\frac{1}{z}$ in the left integral and $u = -1-\frac{1}{z}$ in the right integral, and making use of the relation $\theta_3(z+1) = (-iz)^{-\frac{1}{2}} \theta_2\left(-\frac{1}{z}\right)$, we get
$$h(x) = e^{-i\frac{\pi}{4}}\left(\int_{i\infty}^{\frac{1}{1-b}} ((x+1)z-1)^{-\frac{3}{2}} \theta_2(z) dz + \int_{-\frac{1}{b+1}}^{i\infty} ((x-1)z-1)^{-\frac{3}{2}} \theta_2(z) dz\right)$$
Then we integrate by parts and find
$$h(x) = 4 \frac{\theta_3(b) (x+b)^{\frac{1}{2}}}{1-x^2}+e^{-i\frac{\pi}{4}}\left(\int_{i\infty}^{\frac{1}{1-b}} \frac{2}{x+1} ((x+1)z-1)^{-\frac{1}{2}} \theta_2'(z) dz + \int_{-\frac{1}{b+1}}^{i\infty} \frac{2}{x-1} ((x-1)z-1)^{-\frac{1}{2}} \theta_2'(z) dz\right)$$
Letting $b \to -x$, the boundary term goes to $0$. Then writing $z = w+\frac{1}{1+x}$ in the left integral and $z = w+\frac{1}{x-1}$ in the right integral, we obtain
$$h(x) = 2e^{-i\frac{\pi}{4}} \left(\frac{1}{x+1}\int_{i\infty}^0 ((x+1)w)^{-\frac{1}{2}} \theta_2'\left(w+\frac{1}{x+1}\right) dw + \right.\\
\left.\frac{1}{x-1} \int_0^{i\infty} ((x-1)w)^{-\frac{1}{2}} \theta_2'\left(w+\frac{1}{x-1}\right) dw\right)$$
where $\arg((x+1)w), \arg((x-1)w) \in \left(\frac{3\pi}{2}, 2\pi\right)$. Simplifying slightly, it becomes
$$4\left(-(x+1)^{-\frac{3}{2}} \int_0^\infty t^{-\frac{1}{2}} \sum_{n=1}^\infty e^{i\pi \left(n-\frac{1}{2}\right)^2\left(it+\frac{1}{x+1}\right)}i\pi\left(n-\frac{1}{2}\right)^2 dt\right.\\
\left.+(x-1)^{-\frac{3}{2}} \int_0^\infty t^{-\frac{1}{2}} \sum_{n=1}^\infty e^{i\pi \left(n-\frac{1}{2}\right)^2\left(it+\frac{1}{x-1}\right)}i\pi \left(n-\frac{1}{2}\right)^2 dt\right)$$
By swapping integral and sums and simplifying, we obtain finally that
$$h(x) = 4\pi i\left(-(x+1)^{-\frac{3}{2}} f\left(\frac{1}{x+1}\right)+(x-1)^{-\frac{3}{2}} f\left(\frac{1}{x-1}\right)\right)
$$
where we take $\arg(x+1), \arg(x-1) \in \left(\pi, \frac{3\pi}{2}\right)$ when $\operatorname{Im}(x) < 0, \operatorname{Re}(x) < -1$. By analytic continuation, this holds whenever $x$ is in the sheet $L_1$ and we take $\arg(x+1), \arg(x-1) \in (\pi, 2\pi)$.
Making use of the relation $h(x) = (-ix)^{-\frac{3}{2}} h\left(-\frac{1}{x}\right)$ allows us to derive from this a similar expression for $h$ on the sheet $L_1'$, which is
$$
h(x) = 4\pi i\left((x+1)^{-\frac{3}{2}} f\left(\frac{1}{x+1}\right)+(x-1)^{-\frac{3}{2}}f\left(\frac{1}{x-1}\right)\right)$$
where $\arg(x-1), \arg(x+1) \in (\pi, 2\pi)$.
Putting $z = \frac{1}{x-1}$ and solving for $f(z)$ in the first of these two expressions, we obtain
$$f(z) = (2z+1)^{-\frac{3}{2}} f\left(\frac{z}{2z+1}\right) + \frac{z^{-\frac{3}{2}}}{4\pi i} h\left(\frac{1}{z}+1\right)$$
when $\operatorname{Im}(z) > 0$, where $\arg(2z+1) \in (0, \pi)$ and $\arg(z) \in (-2\pi, -\pi)$. If we fix $b$ at say $-1+i$, we have by the same reasoning as before when $\operatorname{Re}(x) > 1$ that
$$
h(x) = \int_1^b (u+x)^{-\frac{3}{2}} \theta_3(u) du + e^{-i\frac{\pi}{4}}\int_{-\frac{1}{b+1}}^{i\infty} ((x-1)u-1)^{-\frac{3}{2}} \theta_2(u) du
$$
Clearly the first term is bounded as $x \to 1$. The second term, after replacing $x$ with $\frac{1}{z}+1$ and multiplying by $z^{-\frac{3}{2}}$, is
$$
e^{-i\frac{\pi}{4}} \int_{-\frac{1}{b+1}}^{i\infty} (u-z)^{-\frac{3}{2}} \theta_2(u) du
$$
which clearly goes to $0$ as $\operatorname{Re}(z) \to \infty$. It doesn't go to $0$ as $\operatorname{Re}(z) \to -\infty$ because the contour of integration cannot stay fixed. In order to fix the contour of integration when $\operatorname{Re}(z) \to -\infty$, we solve for $f(z)$ in the second expression for $h(z)$ on the sheet $L_1'$, and we obtain
$$
f(z) = -(2z+1)^{-\frac{3}{2}} f\left(\frac{z}{2z+1}\right) + \frac{z^{-\frac{3}{2}}}{4\pi i} h\left(\frac{1}{z}+1\right)
$$
where now that $h$ is evaluated on the sheet $L_1'$, we may show as before that the second term goes to $0$ when $\operatorname{Re}(z) \to -\infty$.
