Let $$f:=\frac{d\mu}{d\nu}.$$ Then
$$f\in L^{\infty}_{\nu,loc}(X)\iff\text{$\forall$ compact $K\subseteq X$ $\exists$ $c_K\in(0,\infty)$ $\forall$ Borel $A$ we have $\mu(A\cap K)\le c_K\nu(A\cap K)$.}$$

Indeed, for the $\Rightarrow$ implication, take any compact $K\subseteq X$. Then $\exists$ $c_K\in(0,\infty)$ such that $f\le c_K$ $\nu$-a.e. on $K$. So, for any
Borel $A$ we have
$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\le c_K\nu(A\cap K),$$
as desired.

Vice versa, for the $\Leftarrow$ implication, take any compact $K\subseteq X$ and suppose that $\mu(A\cap K)\le c_K\nu(A\cap K)$ for some $c_K\in(0,\infty)$ and all Borel $A$. Let now $A:=f^{-1}((c_K,\infty))$, so that $f>c_K$ on $A$. Then
$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\ge c_K\nu(A\cap K),$$
and the latter inequality is strict (and hence contradicts condition $\mu(A\cap K)\le c_K\nu(A\cap K)$) if $\nu(A\cap K)>0$. So, $\nu(A\cap K)=0$, that is, $f\le c_K$ $\nu$-a.e. on $K$, as desired.

Similarly, for any compact $K\subseteq X$,
$$\operatorname{esssup}_Kf>0\iff \text{ $\exists$ $b_K\in(0,\infty)$ $\exists$ Borel $A$ such that $\mu(A\cap K)\ge b_K\nu(A\cap K)$.}$$

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