Let $X$ be a compact metric space and $f:X\to Y$ a quotient map, where $Y$ is Hausdorff. For each $y\in Y$ let $F_y=\{x\in X:f(x)=y\}$, the fibre of $f$ over $y$; $\{F_y:y\in Y\}$ is a partition of $X$ into closed sets. Let $\mathscr{B}$ be a countable base for $X$; without loss of generality we may assume that $\mathscr{B}$ is closed under finite unions.
Let $K\subseteq X$ be closed; then $K$ is compact, so $f[K]$ is compact and hence closed (since $Y$ is Hausdorff). Thus, $f$ is a closed map, and for each closed $K\subseteq X$ the set
$$f^{-1}\big[f[K]\big]=\bigcup_{y\in f[K]}F_y$$
is closed in $X$. It follows that if $U\subseteq X$ is open, then so is
$$\widehat U=\bigcup\{F_y:y\in Y\text{ and }F_y\subseteq U\}\;.$$
In words, the outer saturations of closed subsets of $X$ are closed, and the inner saturations of open subsets of $X$ are open.
Let
$$\mathscr{B}_Y=\left\{f[\widehat B]:B\in\mathscr{B}\right\}\;;$$
$\mathscr{B}_Y$ is a countable family of open subsets of $Y$. Let $y\in Y$, and let $U$ be an open nbhd of $y$ in $Y$. Then $f^{-1}[U]$ is an open nbhd of $F_y$ in $X$. For each $x\in F_y$ there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq f^{-1}[U]$. $\{B_x:x\in F_y\}$ is an open cover of $F_y$, $F_y$ is compact, and $\mathscr{B}$ is closed under finite unions, so there is a $B\in\mathscr{B}$ such that $F_y\subseteq B\subseteq f^{-1}[U]$. Then $F_y\subseteq\widehat B\subseteq f^{-1}[U]$, so $y\in f[\widehat B]\subseteq U$, where $f[\widehat B]\in\mathscr{B}_Y$. Thus, $\mathscr{B}_Y$ is a countable base for the compact Hausdorff space $Y$, which is therefore metrizable.
The answer to the second question is no. Let $X=[0,1]$, $Q=[0,1]\cap\Bbb Q$, and let $Y=X/Q$. If $f$ is the quotient map, it’s not hard to check that $U\subseteq Y$ is open iff $f^{-1}[U]$ is an open nbhd of $Q$ in $[0,1]$. Thus, $Y\setminus\{f(x)\}$ is open in $Y$ for each $x\in X\setminus Q$. If $Y$ were second countable, the singleton $f[Q]$ in $Y$ would therefore be a $G_\delta$ in $Y$, and $Q$ would be a $G_\delta$ in $X$, contradicting the Baire category theorem.