Why the derivatives $f^{(n)}(x)$ of Flat functions grows so fast? (intuition behind)
In this other question I did about Bump functions, other user told in an answer that these kind of functions "tends aggressively fast to zero at the limits of the support", and it keeps in my mind trying to understand what this is supposedly means:
From one side, the functions speed, acceleration, and any other derivatives are bounded since these functions are smooth $C_{c}^{\infty}$, which implies that is completely the opposite was being proposed, the speed and acceleration is finite as any other derivatives, and they looks being well-behaved functions as you expect for something smooth (at least for me, since I relate smooth with polynomials).
But on the other hand, this user and his answer has a point: I have noticed that as I take increasing order derivatives of Flat functions like Bump functions, their higher derivatives maximum values start increasing very fast $\|f^{(n)}(x)\|_{\infty}$ as $n$ grows, and as I understood the formulas, I don't get the intuition behind it.
For me its very counter-intuitive that something that is becoming a constant with bounded speed ($\|f^{(1)}(x)\|_{\infty} \ll \infty$) somehow have higher derivatives are kind of blowing up, which is even more weird in the case of Bump functions since they are smooth $C_{c}^{\infty}$. And with becoming a constant, I mean that for continuously becoming flat all the derivatives must becomes zero at these special inflection points$x^*$ since $\frac{d^n}{dx^n}f(x)\Biggr|_{x\to {x^*}^{\pm}}=0,\,n\geq 1,\, n\in \mathbb{Z}$.
As example, for polynomials as I increase the order of their derivatives, they start to approach to a straight line as their constituent polynomial reduce is order (at least until you match the order of the polynomial), but instead in Flat functions they start to blowing up similarly to what happens when taking $n^{th}$-differences among samples of a Brownian motion, which are in converse very irregular nowhere-differentiable continuous functions (they are actually example of pathological functions, but also as are Flat functions since they aren't analytical in their full domains - check Non-analytic smooth function).
To use a simple example as common ground for the answers let think of this one:$$f(x) = \begin{cases} 0,\quad x\leq -1;\\0,\quad x\geq 1;\\1,\quad x=0;\\\dfrac{1}{1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)},\quad\text{otherwise;}\end{cases}$$You can check the plot of $f(x)$ in Desmos.
Their first derivatives'uniform norms are:$$\|f(x)\|_{\infty} = 1;\quad \|f'(x)\|_{\infty} = 2;\quad \|f''(x)\|_{\infty} \approx 9.84;\quad \|f^{(3)}(x)\|_{\infty} \approx 110.56;\quad \|f^{(4)}(x)\|_{\infty} \approx 2,280.4$$As you could check in Desmos. As you could see they grow very fast, I tried to make a plot $n\,\text{vs}\|f^{(n)}\|_{\infty}$ and its becomes somehow linear only when applying $\ln(1+\ln(1+\|f^{(n)}\|_{\infty}))$ so I think it grows even greater than exponential, all this even when, at least visually, they look quite smooth and even comparable with a traditional quadratic decay (plot), so behaving as normal as traditional polynomials do:
So I would like to know What is actually happening here on this specific points the functions become flat, since I have arguments, I think, opposite from each other being true both at the same time, so I hope you could help me un improve my intuition about this situation.
PS: I use a lot of non-accurate meanings of some terms which I have indicated with italics, this as I am trying to improve the intuition behind and not just a formal formulation. I hope you answers also could focus in the intuition (that why I used the soft-question tag), but also formal demonstration are welcomed: my knowledge in modern abstract math is very limited (for not saying null), so please try to kept the answers at a level of undergraduate engineering courses if possible. Thanks you beforehand.
Added later
After the comment by @MartinR, I checked where the peaks of the higher order derivatives are located (I scaled them in order to be comparable) Desmos, and they start to approach the points when the flat function became zero, so there is almost no amplitude there and the slope is even lower than the slope of the function $y(x)=x$:
Since the speed is almost zero there, I am tempted to say that the situation is more like a mathematical artifactthan saying "tends aggressively fast to zero" near the flat points, which could be seen because $f'(x) \approx 0$ near that neighborhood: Would you agree with that? Or are there some important technical thing I am missing with this interpretation? (please explain it)