$\newcommand{\ep}{\varepsilon}$
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.