Fix a Schwartz function $g_0 : {\bf R} \to {\bf C}$,
say $g_0(x) = e^{-\pi x^2}$ (which looks like a natural choice in this context).

For $x_1,x_2 \in \bf R$ define $g(x_1,x_2)$ by the absolutely convergent sum
$$
g(x_1,x_2) = \sum_{n=-\infty}^\infty g_0(x_2+n) e^{2\pi i n x_1}.
$$
Clearly $g(x_1,x_2) = g(x_1+1,x_2)$ for all $x_1,x_2$.  Also
$$
g(x_1,x_2) = \sum_{n=-\infty}^\infty g_0(x_2+n+1) e^{2\pi i {n+1} x_1}
= e^{2\pi i x_1} \sum_{n=-\infty}^\infty g_0(x_2+n+1) e^{2\pi i n x_1}
= e^{2\pi i x_1} g(x_1, x_2 + 1).
$$
Finally define
$$
f(x_1,x_2) = e^{\pi i x_1 x_2} g(x_1,x_2).
$$
Then $f$ is a smooth function satisfies the required identities
$$
f(x_1,x_2) = e^{-i\pi x_2} f(x_1+1,x_2), \quad 
f(x_1,x_2) = e^{i\pi x_1} f(x_1,x_2+1)
$$
for all real $x_1,x_2$.

This construction was surmised by working backwards, observing that
if $f$ satisfies the required quasiperiodicity then $g$ is periodic in $x_1$
and thus has a Fourier series 
$\sum_{n=-\infty}^\infty g_n(x_2) e^{2\pi i n x_1}$,
and then the identity $g(x_1,x_2) = e^{2\pi i x_1} g(x_1, x_2 + 1)$
yields $g_n(x_2+1) = g_{n+1}(x_2)$, whence $g_n(x_2) = g_0(x_2+n)$ for each $n$.