This is a little too long for a comment, but I hope it might help.
Note that the function $$f_a(t,x) = \sum_{m=-\infty}^\infty (-1)^m k_a(t+mx)$$ $$k_a(t) = (t+a t^3) e^{-t^2} $$ for $a \geq 0$ should satisfy all the requirements you ask - $k_a(t)$ can be written as $h(t)e^{-g(t)}$ where $h(t)$ is odd, has a simple zero at $t=0$, and is positive on the positive real line, and $g(t)$ is even.
In the following, limit oneself to the case of $x=1$.
Notice that $$l_1(t) = \sum_{m=-\infty}^\infty (-1)^m (t+m) e^{-(t+m)^2}$$ is positive on the interval $t \in (0,1)$ by inspection in Mathematica.
Notice also that $$l_3(t) = \sum_{m=-\infty}^\infty (-1)^m (t+m)^3 e^{-(t+m)^2}$$ is negative on the same interval by inspection in Mathematica.
Note that $f_a(t,1) = l_1(t)+a l_3(t)$. Let $a^*$ be the value of $a$ for which $a<a^*$ has $f_a(t,1)$ being positive everywhere on the interval $t \in (0,1)$ and for which $a>a^*$ has $f_a(t,1)$ negative on at least part of this interval.
It's tempting to me that there's some small epsilon for which $f_{a^*+\epsilon}(t,1)$ has at least one simple zero on the interval $t\in(0,1)$ - I think the only way to avoid this is if $f_3(t) = -\frac{1}{a^*}f_1(t)$ (so that one goes abruptly from being positive on the interval $t \in (0,1)$ to negative on the interval), which would be very surprising to me.
Edit: Curiously, $\sum_{m=-\infty}^\infty (-1)^m (t+m) e^{-(t+m)^2}$ and $\sum_{m=-\infty}^\infty (-1)^m (t+m)^3 e^{-(t+m)^2}$ are very nearly (numerically) proportional for $t \in (0,1)$, with a proportionality constant of $\approx -1.0337$. Perhaps this points to the origin of the difficulty in seeing a simple zero.
This (approximate?) proportionality weakens as the argument in the exponential grows, and I think I've found my desired counterexample:
$$f(t,x) = \sum_{m=-\infty}^\infty (-1)^m k(t+mx)$$ $$k(t) = (t+38 t^3) e^{-1.6 t^2}$$
However, this is merely numerical and perhaps not entirely convincing - I'd welcome others to probe this function.