Let $\{p_n(\cdot)\}_{n \in \mathbb{N}}$ be a sequence of polynomials on $[0,1]$, e.g., $p_n(x) = \sum_{i=3}^{d_n} p_{i,n} x^i$, where $d_n$ is the maximal degree which is increasing (at most linearly) in $n$, and $p_{i,n}$ are non-negative coefficients such that $\sum_{i=3}^{d_n} p_{i,n} = 1$. For each $n$ further let $p_n(0)=0$ and $p_n(1)=1$, and assume that $\sum_{i=3}^{d_n} i p_{i,n}$ diverge to $+\infty$ as $n \to +\infty$. The second-order derivative of $p_n(\cdot)$ is given by \begin{equation} p^{''} _n (x) = \sum_{i=3}^{d_n} i(i-1) p_{i,n} x^{i-2}. \end{equation} Take for example $d_n = n$ and $p_{n, n} = 1$ so that $p_n(x)=x^{n}$. Then $p^{''} _n (x) = n(n-1)x^{n-2}$. For fixed $ 0 < x < 1$, this converges to $0$ as $n$ tends to infinity.
Does it hold in general that, for fixed $0 < x < 1$, the sequence of second-order derivatives $\{p_n ^{''}(x)\}_{n \in \mathbb{N}}$ is bounded from above by a finite constant?
Assuming you allow your constant to depend on $x$, the answer is yes. Indeed, fix $x \in (0,1)$ and let $$C_x = \sup_{i \ge 3} i(i-1)x^{i-2}$$ (which must be finite, since $\lim_{i \to \infty} i(i-1)x^{i-2} = 0$). Then, $$\sum_{i=3}^{d_n} i(i-1)p_{i,n}x^{i-2} \leq \sum_{i=3}^{d_n} C_x p_{i,n} = C_x.$$
On the other hand, $C_x \to \infty$ as $x \to 1$ (can you see why?), so we cannot get a uniform bound for all $x$.