$\newcommand{\tla}{\tilde\lambda}\newcommand{\Ga}{\Gamma}$By Definition 4.1 in the paper by Bobkov and Madiman (BM), a positive random variable (r.v.) $\xi$ has a log-concave distribution of order $p\ge1$ if the pdf $f$ of $\xi$ is such that
\begin{equation*}
f(x) = x^{p-1}g(x) \tag{1}\label{1}
\end{equation*}
for $x > 0$, where the function $g$ is log-concave on $(0,\infty)$.
Corollary 3.2 in BM states that, if a positive r.v. $\eta$ has a log-concave pdf, then
\begin{equation*}
\tla_p:=\frac{E\eta^p}{\Ga(p+1)}
\end{equation*}
is log concave in $p\ge0$. It follows then that $\tla_{p+1}\tla_{p-1}\le\tla_p^2$ for $p\ge1$, that is,
\begin{equation*}
E\eta^{p+1}\,E\eta^{p-1}\le\frac{p+1}p\,(E\eta^p)^2. \tag{2}\label{2}
\end{equation*}
Suppose now that a positive r.v. $\xi$ indeed has a log-concave distribution of order $p\ge1$, so that \eqref{1} holds for some log-concave function $g$ and all $x > 0$. Let
\begin{equation*}
h:=g/c,
\end{equation*}
where $c:=\int_0^\infty g$, so that $h$ is a log concave pdf on $(0,\infty)$. Let then $\eta$ be a r.v. with pdf $g$, so that \eqref{2} holds and
\begin{equation*}
E\xi^k=\int_0^\infty x^k f(x)\,dx=\int_0^\infty x^{k+p-1}g(x)\,dx=c\,E\eta^{k+p-1} \tag{3}\label{3}
\end{equation*}
for all $k\in\{0,1,\dots\}$.
Using \eqref{3} with $k=0,1,2$, we rewrite \eqref{2} as
\begin{equation*}
E\xi^2\le\frac{p+1}p\,(E\xi)^2,
\end{equation*}
which can be further rewritten as
$$Var\,\xi\le\frac1p\,(E\xi)^2,$$
as desired.
