$\newcommand\De\Delta\newcommand{\Ga}{\Gamma}\newcommand\ga{\gamma}$Let $X_j:=g(U_j)$, where $g:=|f|$. Then the $X_i$'s are i.i.d. random variables and $g$ is a $1$-Lipschitz function. We have
$$M:=\max_{[0,1]}g=g(u)$$
for some $u\in[0,1]$.
For any real $h>0$,
\begin{align*}
P(\De_n(f)>h)&=P(M-\max_1^n X_j>h) \\
&=P(\max_1^n X_j<M-h) \\
&=P(X_1<M-h)^n \\
&=P(g(U_1)<g(u)-h)^n \\
&=P(g(u)-g(U_1)>h)^n \\
&\le P(|U_1-u|>h)^n \tag{1}\label{1} \\
&\le(1-h)^n\le e^{-nh};
\end{align*}
the $\le$ inequality in \eqref{1} holds because $g$ is $1$-Lipschitz.
So, $\De_n(f)$ is stochastically dominated by an exponentially distributed random variable with mean $1/n$, which confirms your expectation. In particular, $E\De_n(f)\le1/n$.
Consider now
\begin{equation*}
\De_n(F):=\sup_{f\in F}\De_n(f),
\end{equation*}
where $F$ is the set of all $1$-Lipschitz functions on $[0,1]$. Let us show that
\begin{equation*}
E\De_n(F)\sim\frac{\ln n}{2n} \tag{2}\label{2}
\end{equation*}
and
\begin{equation*}
\De_n(F)\Big/\frac{\ln n}{2n}\to1 \tag{3}\label{3}
\end{equation*}
in probability (as $n\to\infty$).
Indeed, it is easy to see that
\begin{equation*}
\De_n(F)=\frac12\,\max(2G_1,G_2,\dots,G_n,2G_{n+1})
=\frac{M_n}2+L_n, \tag{4}\label{4}
\end{equation*}
where
\begin{equation*}
M_n:=\max(G_1,G_2,\dots,G_n,G_{n+1}),
\end{equation*}
\begin{equation*}
0\le L_n\le\frac12\,\max(G_1,G_{n+1}), \tag{5}\label{5}
\end{equation*}
\begin{equation*}
G_i:=U_{n:i}-U_{n:i-1}
\end{equation*}
for $i=1,\dots,n+1$; $U_{n:1}\le\dots\le U_{n:n}$ are the order statistics for the $U_j$'s; $U_{n:0}:=0$; and $U_{n:n+1}:=1$.
A result by Fisher 1929 (see e.g. formula (1.7)) is that for $x\in[0,1]$ we have
\begin{equation*}\label{eq:fisher}
P(M_n>x)=\sum_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}j(1-jx)_+^n,
\end{equation*}
where $u_+:=\max(0,u)$. So,
\begin{equation*}
EM_n=\int_0^1dx\,P(M_n>x)
=\sum_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}j\int_0^1dx\,(1-jx)_+^n
=\frac{\psi(n+2)+\ga}{n+1}\sim\frac{\ln n}n, \tag{6}\label{6}
\end{equation*}
where $\psi:=\Ga'/\Ga$ and $\ga=0.577\dots$ is Euler's gamma; these results for $EM_n$ were also given at the end of Section 1 of the linked paper.
Writing $EM_n^2=\int_0^1dx\,2x\,P(M_n>x)$, we similarly get
\begin{equation*}
Var\,M_n\sim\frac{\pi^2}{6n^2}.
\end{equation*}
So, by \eqref{6} and Chebyshev's inequality,
\begin{equation*}
M_n\Big/\frac{\ln n}n\to1 \tag{7}\label{7}
\end{equation*}
in probability.
In view of \eqref{5}, it is easy to see that $EL_n=O(1/n)$ and hence $L_n/\frac{\ln n}n\to0$ in probability.
Now \eqref{2} and \eqref{3} follow from \eqref{4}, \eqref{6}, and \eqref{7}. $\quad\Box$
From the first equality in \eqref{4} and formula (2.2) of the linked paper, one can get an exact expression for the cdf of $\De_n(F)$: for all real $x\ge0$,
\begin{equation*}
P(2\De_n(F)\le x)\\
=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}k
[(1-kx)_+^n-2(1-(k+\tfrac12)x)_+^n+(1-(k+1)x)_+^n].
\end{equation*}
This yields an exact expression for $E\De_n(F)$:
\begin{equation*}
E\De_n(F)=\frac1{2(n+1)}\Big(\psi(n)+\ga+\frac{2\sqrt{\pi}\,\Ga (n)}{\Ga(n+1/2)} -\frac1n\Big),
\end{equation*}
which, in turn, again yields \eqref{2}.
