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, especially given that it appears quite close to zero everywhere on the interval.
Edit 2:
We can reach a strong understanding of the behavior of these series by noting the following:
For $$f(t,x) = \sum_{m=-\infty}^\infty (-1)^m k(t+mx),$$ if one Fourier expands $$f(t,x) = \sum_{n=-\infty}^\infty a_n(x) e^{2 n \pi i \frac{t}{2x}},$$ one finds that the Fourier coefficient $a_n(x)$ is directly related to the Fourier transform of $k$: $$a_{2n}(x)=0, \,\,\,\, a_{2n+1}(x) = \frac{1}{x}\hat{k}\left(\frac{2n+1}{2x}\right)$$ where $$\hat{k}(\omega) = \int_{-\infty}^\infty e^{-2\pi i \omega t} k(t) dt$$
This links the properties of the $f(t,x)$ series to the Fourier transformation of $k(t)$. I think there is a wealth of information to be gleaned from this. In the following, I'll largely consider the regime of small $x$.
The functions considered in the question and answer are smooth, so their Fourier coefficients decay very rapidly. For small $x$, this means that the dominant contributions to the Fourier series for $f(x,t)$ comes from $a_1(x)$ and $a_{-1}(x)$, with subleading contributions from $a_3(x)$ and $a_{-3}(x)$. Since $k(t)$ is taken to be odd for this question, this gives a Fourier sine series that is dominated by $\sin(2 \pi \frac{t}{2x})$ with a subleading $\sin(2 \pi \frac{3 t}{2x})$ for small $x$.
This explains some of the curious features of my example with $k(t) =(t+38 t^3) e^{-1.6 t^2}$ - the numbers of $38$ and $1.6$ here incidentally work out to nearly cancel the $\sin(2 \pi \frac{t}{2x}) = \sin(\pi t)$ contribution for $x=1$, leaving the subleading small $\sin(2 \pi \frac{3t}{2x}) = \sin(3 \pi t)$ contribution. This behavior is indeed confirmed in the plot above. I had found the numbers $38$ and $1.6$ by hand in Mathematica, but using the Fourier transforms of $t e^{-t^2}$ and $t^3 e^{-t^2}$ should allow one to get a nice analytic handle on choices of coefficients to cancel out the leading behavior of $\sin(2 \pi \frac{t}{2x})$ at small $x$, and to similarly estimate the relative size of the subleading $\sin(2 \pi \frac{3t}{2x})$ contribution.
