Unfortunately I haven't been able to find an intuitive explanation for this behavior or even an alternative proof. But I've played a bit with your proof and I'm posting my version here in the hope that it might help someone else to find an explanation.

Definition

1. Let $\mathcal{S} = \{f \in C^\infty([-1, 1])\colon \forall n\in \mathbb{N}\colon f^{(n)}(-1) = f^{(n)}(1) = 0 \}$.
2. For $n\in\mathbb{N}$ and $\alpha\in\mathbb{R}\setminus[-1, 1]$ let $D_\alpha^n = \frac{1}{(x + \alpha)^{n-1}}\frac{d}{dx}(x+\alpha)^n$.
3. For $f\in\mathcal{S}$ let $Z(f)\in \mathbb{N}\cup \{\infty\}$ be the number of zeros of $f$ in $(-1, 1)$.

Lemma

1. $Z(D^n_\alpha f) \ge 1 + Z(f)$.
2. $\left((x+\alpha)f\right)^{(n)} = D_\alpha^n f^{(n-1)}$. $\square$

Proof

1. $f$ (and thus $(x+\alpha)^n f$) has $Z(f)+2$ zeros on $[-1, 1]$. By Rolle's theorem, $D_\alpha^n f$ has at least $Z(f)+1$ zeros in $(-1, 1)$.
2. $$((x+\alpha)f)^{(n)} = nf^{(n-1)} + (x + \alpha)f^{(n)} = \frac{1}{(x+\alpha)^{n-1}}\frac{d}{dx}\left((x+\alpha)^n f^{(n-1)}\right) = D_\alpha^n f$$

Theorem

For all $f\in\mathcal{S}$ there is an $n\in\mathbb{N}$ such that $Z(f^{(n)})> n + 1$.

Proof

If $f$ has a zero in $(-1, 1)$, we are done. Otherwise assume that $f>0$ in $(-1, 1)$ and let $\alpha = \frac{6}{5}$ and $\beta = \frac{4}{5}$. Note that $|\alpha^2 - \beta^2|<1$. Let $g_k = \frac{f}{(x^2 - \alpha^2)^{2k}}$. Then

• $g_k(-1) = 0$,
• $g_k(-\beta) \to \infty$,
• $g_k(0) \to 0$,
• $g_k(\beta) \to \infty$, and
• $g_k(1) = 0$

for $k\to\infty$. Hence, for $k$ large enough, $g_k$ has at least two local maxima and one local minimum in $(-1, 1)$. Thus, $Z(g_k') > 2$.

By factoring $x^2 - \alpha^2 = (x-\alpha)(x+\alpha)$ and applying the lemma $4k$ times, we see $$f^{(4k+1)} = ((x^2 - \alpha^2)^{2k}g_k)^{(4k+1)} = D_{-\alpha}^{4k+1}D_\alpha^{4k}\cdots D_{-\alpha}^{3}D_\alpha^{2}g_k'.$$ Hence, $Z(f^{(4k+1)}) \ge 4k + Z(g_k') > 4k + 2$. $\square$

Unfortunately I haven't been able to find an intuitive explanation for this behavior or even an alternative proof. But I've played a bit with your proof and I'm posting my version here in the hope that it might help someone else to find an explanation.

Definition

1. Let $\mathcal{S} = \{f \in C^\infty([-1, 1])\colon \forall n\in \mathbb{N}\colon f^{(n)}(-1) = f^{(n)}(1) = 0 \}$.
2. For $n\in\mathbb{N}$ and $\alpha\in\mathbb{R}\setminus[-1, 1]$ let $D_\alpha^n = \frac{1}{(x + \alpha)^{n-1}}\frac{d}{dx}(x+\alpha)^n$.
3. For $f\in\mathcal{S}$ let $Z(f)\in \mathbb{N}\cup \{\infty\}$ be the number of zeros of $f$ in $(-1, 1)$.

Lemma

1. $Z(D^n_\alpha f) \ge 1 + Z(f)$.
2. $\left((x+\alpha)f\right)^{(n)} = D_\alpha^n f^{(n-1)}$.

Proof

1. $f$ (and thus $(x+\alpha)^n f$) has $Z(f)+2$ zeros on $[-1, 1]$. By Rolle's theorem, $D_\alpha^n f$ has at least $Z(f)+1$ zeros in $(-1, 1)$.
2. $$((x+\alpha)f)^{(n)} = nf^{(n-1)} + (x + \alpha)f^{(n)} = \frac{1}{(x+\alpha)^{n-1}}\frac{d}{dx}\left((x+\alpha)^n f^{(n-1)}\right) = D_\alpha^n f^{(n-1)}$$ $\square$

Theorem

For all $f\in\mathcal{S}$ there is an $n\in\mathbb{N}$ such that $Z(f^{(n)})> n + 1$.

Proof

If $f$ has a zero in $(-1, 1)$, we are done. Otherwise assume that $f>0$ in $(-1, 1)$ and let $\alpha = \frac{6}{5}$ and $\beta = \frac{4}{5}$. Note that $|\alpha^2 - \beta^2|<1$. Let $g_k = \frac{f}{(x^2 - \alpha^2)^{2k}}$. Then

• $g_k(-1) = 0$,
• $g_k(-\beta) \to \infty$,
• $g_k(0) \to 0$,
• $g_k(\beta) \to \infty$, and
• $g_k(1) = 0$

for $k\to\infty$. Hence, for $k$ large enough, $g_k$ has at least two local maxima and one local minimum in $(-1, 1)$. Thus, $Z(g_k') > 2$.

By factoring $x^2 - \alpha^2 = (x-\alpha)(x+\alpha)$ and applying the lemma $4k$ times, we see $$f^{(4k+1)} = ((x^2 - \alpha^2)^{2k}g_k)^{(4k+1)} = D_{-\alpha}^{4k+1}D_\alpha^{4k}\cdots D_{-\alpha}^{3}D_\alpha^{2}g_k'.$$ Hence, $Z(f^{(4k+1)}) \ge 4k + Z(g_k') > 4k + 2$. $\square$