Let $$s_q(N):=\sum_{n=0}^N f_q(n),\quad t_q(N):=\sum_{n=1}^N g_q(n),$$ where $$f_q(n):=(4n+1)q^{(4n+1)^2/4},\quad g_q(n):=(4n-1)q^{(4n-1)^2/4}.$$ We want to show that $$s_q(\infty)-t_q(\infty)\overset{\text{(?)}}\ge1/10 \tag{1}$$ for all $$q\in[e^{-2},e^{-1/2}]. \tag{2} $$
For such $q$, $g_q(n)$ is decreasing in $n\ge1$ and increasing in $q$, and hence
$$t_q(\infty)-t_q(2)
=\sum_{n=3}^\infty g_q(n) \\
<\int_2^\infty g_{e^{-1/2}}(u)\,du=e^{-49/8}.$$
So, for $q$ as in (2),
$$
\begin{aligned}
&s_q(\infty)-t_q(\infty) \\
&>s_q(2)-t_q(2)-e^{-49/8} \\
&=h(q):=9 q^{81/4}-7 q^{49/4}+5 q^{25/4}-3 q^{9/4}+q^{1/4}-e^{-49/8} \\
&\ge h(e^{-1/2})>1/10;
\end{aligned}$$
the penultimate inequality, $h(q)\ge h(e^{-1/2})$, in the above multiline display is easy to prove, since $h(q)$ is a simple polynomial in $q^{1/4}$.
So, (1) indeed holds for all $q$ as in (2).