$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\De{\Delta}$We have
$$\frac{f''(x)}n=\sum_{k=0}^{n-1} h(k)\binom{n-1}k x^k(1-x)^{n-1-k}, \tag{10}\label{10}$$
where
$$h(y):=y g(y-1)-2 (y+1) g(y)+(y+2) g(y+1)$$
and $g(y):=\ln(1+\max(0,y))$, so that
$$h(k)\ge0 \tag{20}\label{20}$$
for all integers $k\ge0$.
So, $f''\ge0$ on $[0,1]$. $\quad\Box$
Clearly, this argument will work for any function $g$ such that $k g(k-1)$ is convex in integral $k\ge0$.
Details on \eqref{10}: For all $x\in[0,1]$,
$$f(x)=x(B_n g)(x),$$
where
$$(B_n g)(x):=\sum_{k\in\Z} g(k)\binom nk x^k(1-x)^{n-k};$$
of course, $\binom nk=0$ and hence $g(k)\binom nk=0$ unless $k\in\{0,\dots,n\}$.
Using the identities $\binom nk k=n\binom{n-1}{k-1}$ and $\binom n{n-k}(n-k) k=n\binom{n-1}k$, for any function $g\colon\R\to\R$ and any natural $n$ we have
\begin{align}
(B_n g)'(x)&=\sum_{k\in\Z} g(k)\binom nk kx^{k-1}(1-x)^{n-k} \\
&-\sum_{k\in\Z} g(k)\binom nk(n-k)x^k(1-x)^{n-k-1} \\
&=n\sum_{j\in\Z} g(j+1)\binom{n-1}j x^j(1-x)^{n-1-j} \\
&-n\sum_{k\in\Z} g(k)\binom{n-1}kx^k(1-x)^{n-1-k} \\
&=n\sum_{k\in\Z} [g(k+1)-g(k)]\binom{n-1}k x^k(1-x)^{n-1-k} \\
&=n (B_{n-1}\De g)(x),
\end{align}
where $(\De g)(y):=g(y+1)-g(y)$. So, $(B_n g)''=n(n-1)B_{n-2}\De^2 g$ and hence
\begin{align}
f''(x)&=2(B_n g)'(x)+x(B_n g)''(x) \\
&=2n(B_{n-1}\De g)(x)+n(n-1)x(B_{n-2}\De^2 g)(x) \\
&=2n\sum_{k\in\Z} (\De g)(k)\binom{n-1}k x^k(1-x)^{n-1-k} \\
&+n(n-1)\sum_{j\in\Z} (\De^2 g)(j)\binom{n-2}j x^{j+1}(1-x)^{n-2-j} \\
&=2n\sum_{k\in\Z} (\De g)(k)\binom{n-1}k x^k(1-x)^{n-1-k} \\
&+n\sum_{k\in\Z} (\De^2 g)(k-1)\binom{n-1}k k x^k(1-x)^{n-1-k},
\end{align}
so that \eqref{10} holds; at the last step here, we used the identity $(n-1)\binom{n-2}{k-1}=\binom{n-1}{k} k$.
Details on \eqref{20}: We have $h(0)=2\ln2>0$ and
$$h(y)=y \ln y-2 (y+1) \ln(y+1)+(y+2) \ln(y+2)>0$$
for real $y\ge1$, because $t\ln t$ is convex in $t>0$.
