$\newcommand{\s}{\overset{\text{sign}}=}$Let us first recall some facts about the Bernoulli polynomials $B_k(x)$ and the Bernoulli numbers, which latter will be denoted here by $b_k$:
\begin{equation*}
B_k(x)=\sum_{j=0}^k\binom nk b_{n-k}x^k \tag{1}\label{1}
\end{equation*}
-- see e.g. this;
\begin{equation*}
b_0=1,\ b_1=-1/2,\ b_{2n+1}=0; \tag{2}\label{2}
\end{equation*}
\begin{equation*}
b_{2n}={\frac {(-1)^{n+1}2(2n)!}{(2\pi )^{2n}}}\zeta(2n) \tag{3}\label{3}
\end{equation*}
-- see e.g. this,
so that
\begin{equation*}
b_{2n}\s(-1)^{n+1}; \tag{4}\label{4}
\end{equation*}
\begin{equation*}
B'_k=kB_{k-1} \tag{5}\label{5}
\end{equation*}
-- see e.g. formulas (1.1) and (1.5) in this paper by Leeming.
Everywhere here, by default $k$ and $n$ are positive integers, unless specified otherwise; and $\s$ denotes the equality in sign.
Next -- see e.g. p. 125 in Leeming's paper,
\begin{equation*}
B_{2n+1}(0)=B_{2n+1}(1/2)=0,\quad B_{2n+1}(x)\ne0\text{ for }x\in(0,1/2), \tag{6}\label{6}
\end{equation*}
\begin{equation*}
\exists! x_{2n}\in(0,1/2)\ B_{2n}(x_{2n})=0. \tag{7}\label{7}
\end{equation*}
It follows from \eqref{5}, \eqref{1}, and \eqref{4} that $B'_{2n+1}=(2n+1)B'(0)\s(-1)^{n+1}$ and hence, by \eqref{6},
\begin{equation*}
B_{2n+1}(x)\s(-1)^{n+1}\text{ for }x\in(0,1/2). \tag{8}\label{8}
\end{equation*}
Next, by \eqref{5} and \eqref{8}, for $n\ge1$ we have $B'_{2n}\s B_{2n-1}\s(-1)^n$ (on $(0,1/2)$). Also, $B_{2n}B'_{2n}>0$ on $(x_{2n},1)$ and $B_{2n}B'_{2n}<0$ on $(0,x_{2n})$. So,
\begin{equation*}
B_{2n}\s
\begin{cases}
(-1)^{n+1}&\text{ on }(0,x_{2n}), \\
(-1)^n&\text{ on }(x_{2n},1).
\end{cases}
\tag{9}\label{9}
\end{equation*}
Moreover, it was shown by Ostrowski (see again, e.g., p. 125 in Leeming's paper) that $x_{2n}<x_{2n+2}$ for all $n\ge1$.
So, letting
\begin{equation*}
r_k:=\frac{B_k}{B_{k+2}}. \tag{10}\label{10}
\end{equation*}
we have
\begin{equation*}
r_{2n}
\begin{cases}
>0&\text{ on }(x_{2n},x_{2n+2})\ne\emptyset, \\
<0&\text{ on }(x_{2n+2},1)
\end{cases}
\end{equation*}
and hence
\begin{equation*}
r_{2n}\to
\begin{cases}
\infty &\text{ as }x\uparrow x_{2n+2}, \\
-\infty &\text{ as }x\downarrow x_{2n+2}.
\end{cases}
\tag{10.5}\label{10.5}
\end{equation*}
After these preliminaries, we see that have to show that $|r_{2n+1}|$ is decreasing on $(0,1/2)$.
In view of \eqref{8}, $r_{2n+1}<0$ (everywhere on $(0,1/2)$). So, our task is to show that $r_{2n+1}$ is increasing on $(0,1/2)$.
We will show a bit more:
\begin{equation*}
\begin{aligned}
&\text{If $k$ is odd, then $r_k$ is increasing on $(0,1/2)$.} \\
&\text{If $k$ is even, then $r_k$ is increasing on $(0,x_{k+2})$ and on $(x_{k+2},1/2)$.}
\end{aligned}
\tag{$\dagger$}\label{dagger}
\end{equation*}
Here $x_{k+2}$ is defined by (7), as the only zero of $B_{k+2}$ in the interval $(0,1/2)$.
Note that, in view of \eqref{6}, $r_k$ is continuous on $(0,1/2)$ if $k$ is odd (and, by \eqref{7}, $r_k$ has the only point of discontinuity, at $x_{k+2}$, if $k$ is even). Claim \eqref{dagger} is illustrated below by the graphs $\{(x,r_k(x))\colon0<x<1/2\}$ for $k=3$ (left) and $k=4$ (right, with only part of the graph shown, because of the infinite discontinuity at $x_{k+2}=x_6$):
The proof of \eqref{dagger} will be done by induction on $k$. The induction base, for $k=0$ and $k=1$, is checked easily, since $B_0(x)=1$, $B_1(x)=x-1/2$, $B_2(x)=1/6 - x + x^2$, and $B_3(x)=x/2 - 3 x^2/2 + x^3$.
Suppose now that the statement \eqref{dagger} holds for some integer $k\ge1$. We then have to show that \eqref{dagger} holds with $k+1$ in place of $k$.
This will be accomplished using so-called l'Hospital-type rules for monotonicity (l'H). To use these rules, consider the "derivative ratio" for $r_{k+1}$ (cf. \eqref{10}):
\begin{equation*}
\rho_{k+1}
:=\frac{B'_{k+1}}{B'_{k+3}}=\frac{k+1}{k+3}\frac{B_k}{B_{k+2}}=\frac{k+1}{k+3}\,r_k,
\end{equation*}
by \eqref{5} and \eqref{10},
so that the monotonicity pattern of $\rho_{k+1}$ is the same as that of $r_k$:
\begin{equation*}
\begin{aligned}
&\text{If $k$ is odd, then $\rho_{k+1}$ is increasing on $(0,1/2)$.} \\
&\text{If $k$ is even, then $\rho_{k+1}$ is increasing on $(0,x_{k+2})$ and on $(x_{k+2},1/2)$.}
\end{aligned}
\tag{11}\label{11}
\end{equation*}
Consider first the case when $k=2n$, even. Then, by Proposition 4.1 in the l'H paper, \eqref{11}, and the condition $B_{2n+1}(0)=B_{2n+1}(1/2)=0$ in \eqref{6}, we see that \eqref{dagger} implies that
$r'_{k+1}>0$ on $(0,x_{k+2})$ and on $(x_{k+2},1/2)$, so that the induction step is done in this case.
Consider now the case when $k=2n-1\ge1$, odd. Then it is not true that $B_{k+1}(0)=0$ or $B_{k+1}(1/2)=0$. So, in this case, we have to use so-called general l'Hospital-type rules for monotonicity. More specifically, we will use Table 1.1, with $f=B_{k+1}=B_{2n}$ and $g=B_{k+3}=B_{2n+2}$, so that $r_{k+1}=f/g$ and, by \eqref{5} and \eqref{8}, $g'\s B_{2n+1}\s(-1)^{n+1}$ on the entire interval $(0,1/2)$. So, by \eqref{7}, $gg'\s B_{2n+2}B_{2n+1}\s B_{2n+2}(-1)^{n+1}$, which, by \eqref{9}, is
$<0$ on the interval $(0,x_{2n+2})=(0,x_{k+3})$ and $>0$ on the interval $(x_{2n+2},1/2)=(x_{k+3},1/2)$. Hence, by
lines 3 and 1 of mentioned Table 1.1 and \eqref{11}, we get that $r_{k+1}=r_{2n}$ is
up-down on $(0,x_{k+3})$ (that is, for some $c_1\in[0,x_{k+3}]$ the function $r_{k+1}=r_{2n}$ is increasing on $(0,c_1)$ and decreasing on $(c_1,x_{k+3})$;
down-up on $(x_{k+3},1/2)$ (that is, for some $c_2\in[x_{k+3},1/2]$ the function $r_{k+1}=r_{2n}$ is decreasing on $(x_{k+3},c_2)$ and increasing on $(c_2,1/2)$.
However, in view of \eqref{10.5}, $r_{2n}$ cannot be decreasing in any left neighborhood of $x_{k+3}=x_{2n+2}$, and $r_{2n}$ cannot be decreasing in any right neighborhood of $x_{k+3}=x_{2n+2}$. So, in the above "up-down" and "down-up" items, $c_1=x_{k+3}=c_2$; that is, $r_{k+1}=r_{2n}$ is increasing on $(0,x_{k+3})$ and on $(x_{k+3},1/2)$.
Thus, whether $k$ is even or odd, \eqref{dagger} holds with $k+1$ in place of $k$, which completes the induction step. $\quad\Box$
