When does $G/Z(G)$ actually have trivial center?

The pullback of the center of $G/Z(G)$ is called the "second center of $G$", $Z_2(G)$; it is the second term of the upper central series of $G$, which in turn is intimately connected with the lower central series of $G$ and to the notion of nilpotency.

Thus, $G/Z(G)$ has trivial center if and only if $Z_2(G)=Z(G)$. For a nilpotent group, this happens if and only if $G$ is abelian, in which case $Z(G)=G$. In particular, if $G/Z(G)$ is of prime power order, and not trivial, then $Z_2(G)$ is strictly larger than $Z(G)$.

There are many conditions that guarantee this; for example, if $G=[G,G]$, then $G/Z(G)$ is centerless.

Now, I think you are viewing quotients wrong; we are not "removing" the subgroup, we are trivializing it: we are making everything in the subgroup not matter (rather than "not being there"). If you view $Z(G)$ as the collection of all elements that commute with everything, then $Z_2(G)$ is the collection of all elements that commute with everything up to elements that "don't matter" (namely, the elements of the center, which commute with everything). That is, $g\in Z_2(G)$ if and only if for every $x\in G$ there exists $z\in Z(G)$ such that $gx = xgz$. If you consider elements of $Z(G)$ to "be trivial", "not matter" (which is what you do when you take the quotient), then $g$ "essentially" commutes with $x$.

Certain properties of groups are closed under extension. That is, if $N$ is a normal subgroup of $G$ and both $N$ and $G/N$ have that property, then so does $G$.

So for example, the properties of being a $p$-group for a specific prime $p$, or of being solvable, or of being a torsion group are closed under extension, whereas the properties abelian and nilpotent are not.

For a property that is closed under extension, if there is a largest normal subgroup $N$ of $G$ having that property (which will always be the case for finite groups), then $G/N$ will have only the trivial normal subgroup with that property.

So, for a finite group $G$ and prime $p$, there is a largest normal $p$-subgroup, the so-called $p$-core $O_p(G)$, and $G/O_p(G)$ has trivial $p$-core. Similarly for the largest normal solvable subgroup.

For infinite groups it is more complicated, because there might not be a maximal normal $p$-subgroup or solvable subgroup.