$\newcommand{\si}{\sigma}\newcommand{\F}{\mathcal F}\newcommand{\G}{\mathcal G}\newcommand{\pa}{\parallel}$The first inequality is obviously false in general, e.g. when $A=B$ but $f(A)$ differs from $g(B)=g(A)$ in distribution.
The second inequality is true. A bit more generally, let $\mu$ and $\nu$ be probability measures defined on a $\si$-algebra $\F$. Let $\mu_\G$ and $\nu_\G$ be the restrictions of $\mu$ and $\nu$ to a sub-$\si$-algebra $\G$ of $\F$.
Then
\begin{equation*}
D_{KL}(\mu_\G\pa\nu_\G)\le D_{KL}(\mu\pa\nu). \tag{1}\label{1}
\end{equation*}
Indeed,
\begin{equation*}
D_{KL}(\mu\pa\nu)=\int d\nu\,h\Big(\frac{d\mu}{d\nu}\Big),
\end{equation*}
where $h(u):=u\ln u$ for $u\in(0,\infty)$, with $h(0):=0$ and $h(\infty)=\infty$.
Next, letting $E_\nu(\cdot|\G)$ denote the conditional expectation given $\G$ w.r.t. $\nu$, for any nonnegative $\G$-measurable function $q$ we have
\begin{equation*}
\int d\nu_\G\,q\, E_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)
= \int d\nu\,q\, E_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)\,
=\int d\nu\,q\, \frac{d\mu}{d\nu}\,
= \int d\mu\,q= \int d\mu_\G\,q;
\end{equation*}
the second equality in the latter display holds by the definition of the conditional expectation. So,
\begin{equation*}
\frac{d\mu_\G}{d\nu_\G}=E_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big).
\end{equation*}
So, by Jensen's inequality applied to the convex function $h$,
\begin{multline*}
D_{KL}(\mu_\G\pa\nu_\G)
=\int d\nu\,h\Big(\frac{d\mu_\G}{d\nu_\G}\Big)
=\int d\nu\,h\Big(E_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)\Big) \\
\le \int d\nu\,E_\nu\Big( h\Big(\frac{d\mu}{d\nu}\Big)\Big|\G\Big)
=\int d\nu\,h\Big(\frac{d\mu}{d\nu}\Big)
=D_{KL}(\mu\pa\nu),
\end{multline*}
so that \eqref{1} follows.
Your second inequality now obtains by letting $\mu$ and $\nu$ be the distributions of $A$ and $B$, respectively, over the measurable space $(E_1,\F)$ and letting $\G$ be the sub-$\si$-algebra of $\F$ generated by the measurable map $f$.
$\quad\Box$
