$\newcommand{\ep}{\varepsilon}$ Letting $a_{it}:=f(|t_i|)$ and $b_{it}:=g(|t_i|)$, rewrite your inequality ($\ast$) as \begin{equation} E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big|\le C\,E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big| \end{equation} for any natural $N$ and any set $T$, with the condition that $0\le a_{it}\le b_{it}$ for all $i,t$.
Let now $T:=2^{[N]}$, the power set of the set $[N]:=\{1,\dots,N\}$. Let $b_{it}:=1$
and
\begin{equation}
a_{it}:=1_{i\in t}
\end{equation}
for all $i\in[N]$ and $t\in T=2^{[N]}$. Then
\begin{equation}
E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big|
\ge E\sum_{i=1}^N\max(0,\ep_i)=N/2,
\end{equation}
whereas
\begin{equation}
E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big|
=E\Big|\sum_{i=1}^N\ep_i\Big|\le\sqrt N.
\end{equation}
So, your inequality ($\ast$) will fail to hold for any $C>0$ if $N$ is large enough.