Between the center of a quotient group and the total center

What is sure is that you have $Z(G)/Z(G)\cap H$ is isomorphic to a subgroup of $Z(G/H)$ (see Nicky Hekster's answer for details about this). Apart from this, you cannot say anything in general.

$Z(G/H)$ is trivial whereas $Z(G)$ is not.

Consider a direct product $G=A \times S$ where $S$ is a non-abelian simple group and $A$ is an abelian group. It is not hard to see that $Z(G)=A\times\{1_S\}$. Furthermore, taking $H:=Z(G)$ then $G/H$ is naturally isomorphic to $S$ which has trivial center.

$Z(G/H)$ is equal to $G/H$ whereas $Z(G)$ is trivial.

Consider $S_3$ the symmetric group over $3$ elements. Then $Z(S_3)$ is trivial. Consider $S_3/A_3$ (where $A_3$ is the alternating group) then $S_3/A_3$ is isomorphic to $\mathbb{Z}/2$ and thus is abelian.

Proposition If $H \unlhd G$ with $H \cap G'=1$, then $H \subseteq Z(G)$ and $Z(G/H)=Z(G)/H$.

Proof $[G,H] \subseteq H$ since $H$ is normal, and of course $[G,H] \subseteq G'$. Hence $[G,H]=1$, which is equivalent to $H \subseteq Z(G)$. Now put $Z(G/H)=K/H$. Then $[G,K] \subseteq H$ and of course $[G,K] \subseteq G'$. Hence $[G,K]=1$, that is $K \subseteq Z(G)$. And since $H \subseteq Z(G)$, also $Z(G)/H \subseteq K/H$.