First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$ \sum_{n=-\infty}^\infty e^{-\alpha (x-n)^2}=2{\sqrt{\frac{\pi}{\alpha}}}\sum_{n=0}^\infty e^{-\frac{\pi^2}{\alpha}n^2}\cos2\pi n x,\quad \text{Re}\alpha>0. $$ From this formula one can see that when $\alpha>0$ is small, the Fourier coefficients $a_n=2{\sqrt{\frac{\pi}{\alpha}}}e^{-\frac{\pi^2}{\alpha}n^2}$ of the resulting function decrease rapidly with increasing $n$. When $\alpha=1/2$ one has $$ \frac{a_1}{a_0}=e^{-2\pi^2}\approx 2.7\times 10^{-9}. $$

First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$ \sum_{n=-\infty}^\infty e^{-\alpha (x-n)^2}=2{\sqrt{\frac{\pi}{\alpha}}}\left(1+2\sum_{n=1}^\infty e^{-\frac{\pi^2}{\alpha}n^2}\cos2\pi n x\right),\quad \text{Re}~\alpha>0.\tag{1} $$ From this formula one can see that when $\alpha>0$ is small, the Fourier coefficients $a_n=2{\sqrt{\frac{\pi}{\alpha}}}e^{-\frac{\pi^2}{\alpha}n^2}$ of the resulting function decreases rapidly with increasing $n$. When $\alpha=1/2$ one has $$ \frac{a_1}{a_0}=2e^{-2\pi^2}\approx 5.4\times 10^{-9}. $$

I don't know any conceptual explanation for this in mathematics, but there is such an explanation that comes from physics. Consider a quantum particle with mass $m$ on a ring of radius $a$. We assume the ring is pierced with magnetic flux $\phi$. We want to calculate partition function of this system at temperature $T>0$ in two different ways.

On the one hand, it is known that the energy spectrum of the particle is given by $$ E_n=\frac{\hbar^2}{2m^2}\left(n+\frac{\phi}{\phi_0}\right)^2, $$ where $\phi_0$ is the so called magnetic flux quantum. Then partition function is given as the Gibbs sum $$ Z=\sum_{n=-\infty}^\infty e^{-E_n/T}=\sum_{n=-\infty}^\infty e^{-\frac{\hbar^2}{2m^2T}\left(n+{\phi}/{\phi_0}\right)^2}.\tag{2} $$ Here one immediately recognizes the sum analogous to the LHS of $(1)$.

On the other hand, partition function is related to the trace of the density matrix $\rho_{\theta_1,\theta_2}$, i.e. to $\int\rho_{\theta,\theta}d\theta$. It is possible to calculate the density matrix of this system in imaginary time representation by solving a certain differential equation. Details of this calculation can be found for example in this book, chapter 4.3. Consider first the more simple case of unbounded line $-\infty<\theta<+\infty$; then the answer is $$ \rho_{\theta_1,\theta_2}=Ce^{-\frac{mTa^2(\theta_1-\theta_2)^2}{2\hbar^2}}, $$ where $C$ is some normalization constant. Magnetic flux does not enter this expression because there is not any nontrivial loop in the system. On a ring there is a nontrivial loop, and let $n$ be the winding number. It is known that the partition function is a sum over all homotopy classes multiplied by corresponding phase factor. In this case the phase factor comes from the magnetic flux (Aharonov-Bohm phase) and equals $e^{2\pi in\phi/\phi_0}$. As a result we have \begin{align} Z&=\sum_{n=-\infty}^\infty e^{2\pi in\phi/\phi_0}\int\rho_{\theta,\theta+2\pi n}d\theta\\ &=2\pi C \sum_{n=-\infty}^\infty e^{-\frac{2\pi^2mTa^2}{\hbar^2}n^2+2\pi in\phi/\phi_0}.\tag{3} \end{align} Combining $(2)$ and $(3)$ one obtains the transformation in $(1)$.

One can see from this physical interpretation that when the temperature $T$ increases, then the higher Fourier harmonics decrease. This corresponds to the physical intuition that the higher the temperature the more chaotic the system becomes and the effect of the Aharonov-Bohm phase averages out due to thermal fluctuations.

This is related to the concept of instanton in quantum mechanics. See for example the book Altland, Simons, Condenced matter theory, page 154.

First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$ \sum_{n=-\infty}^\infty e^{-\alpha (x-n)^2}=2{\sqrt{\frac{\pi}{\alpha}}}\sum_{n=0}^\infty e^{-\frac{\pi^2}{\alpha}n^2}\cos2\pi n x,\quad \text{Re}\alpha>0. $$ From this formula one can see that when $\alpha>0$ is small, the Fourier coefficients $a_n=2{\sqrt{\frac{\pi}{\alpha}}}e^{-\frac{\pi^2}{\alpha}n^2}$ of the resulting function decrease rapidly with increasing $n$. When $\alpha=1/2$ one has $$ \frac{a_1}{a_0}=e^{-2\pi^2}\approx 2.7\times 10^{-9}. $$