I'm dealing with a finite $p$-group $G$ whose derived subgroup $G'$ is cyclic. The author defines $C^*\leq G$ as
\begin{equation*} C^*=C_G\big(G'/(G')^{p^2}\big). \end{equation*}
By definition, the centralizer $C_G(S)$ of a subset $S$ of $G$ is the subgroup given by all elements $g$ of $G$ such that $[g,s]=1\, \forall s\in S$. I'm wondering how I can define a subgroup like $C^*$ as I'm being asked to compute the commutator between an element of $G$ and a coset in $G'/(G')^{p^2}$.
The full article can be read here, this centralizer can be found at page 113.
There are two ways to interpret this, both leading to the same set of elements.
One is, as the_fox mentions, to understand this as the set of elements in $G$ that map, under the canonical map $G\to G/(G')^{p^2}$, to the centralizer of $G'/(G')^{p^2}$. This is seen in the definition of the second center, which is sometimes said to be the centralizer in $G$ of $G/Z(G)$ (well, not literally that, as you note).
The other way to interpret this is that $G$ has a natural action on $G'/(G')^{p^2}$ by conjugation, and you are asking for the centralizer in that action, that is, the elements that fix everything: if $G$ is a group acting on the left on $H$, then for $A\leq H$ we let $C_G(A)=\{g\in G\mid {}^ga = a\text{ for all }a\in A\}$
In fact, the two notions coincide. For simplicity, I will let $(G')^{p^2}=N$.
Suppose $g\in G$ is such that $gN\in C_{G/N}(G'/N)$. That means that for every $x\in G'$, $(Nx)(Ng) = (Ng)(Nx)$, or equivalently, that $Ngxg^{-1} = Nx$. This means that ${}^g(Nc) = Nx$ for all $x\in G'$, so $g$ centralizes $G'/N$ under its standard action.
Conversely, if ${}^g(Nx)=Nx$ for all $x\in G'$, then $Ngxg^{-1}N=Nx$, so $(Ng)(Nx)=(Nx)(Ng)$ in $G/N$; that is, $gN\in C_{G/N}(G'/N)$.
So the two interpretations result in the exact same subgroup of $G$. Of course you can replace $G'$ with some other subgroup containing $N$.