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

$\newcommand{\ep}{\varepsilon}$ 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{E\Big(\sum_{i=1}^N\ep_i\Big)^2}=\sqrt N. \end{equation}

So, your inequality ($\ast$) will fail to hold for any $C>0$ if $N>4C^2$.

$\newcommand{\ep}{\varepsilon}$ 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\sup_{t\in T}\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.

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

Under the additional condition that $$|f(x)-f(y)|\le|g(x)-g(y)| \tag{1}$$ for all $x$ and $y$, your desired inequality ($\ast$) holds with $C=1$. This follows from Lemma 5, page 7 (note also the remark "Both definitions agree for function classes which are closed under negation" in line -10 on the same page). The proof of that lemma is short and very clever. I am afraid that without (1) your inequality (*) may fail to hold, even though it may be hard to construct a counterexample, given the presence of the factor $C$.

$\newcommand{\ep}{\varepsilon}$ 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\sup_{t\in T}\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.