Let $G$ be a group and $Z(G)$ be its center. Now let $G_{1}$ be the group $G/Z(G)$ and $Z(G_{1})$ be its center. Inductively let $G_{i}$ be the group $G_{i-1}/Z(G_{i-1})$.
If for some $k$ we have that $G_{k}/Z(G_{k})$ is abelian can we deduce that the original group is nilpotent?
I think that there is a connection with the upper central series of a group defined by the normal subgroups
$$Z_{i}(G)=\{g\in G : [g,h] \in Z_{i-1}(G) \hspace{1mm} \forall h \in G\}$$
but I'm not sure I fully understand it.
You are just defining the quotients of $G$ by its upper central series, but with different notation.
Recall that the upper central series is defined recursively by letting $Z_0(G)=\{e\}$, $Z_1(G)=Z(G)$, and $Z_{i+1}(G)$ to be the subgroup of $G$ such that $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$.
I claim that your definition of $G_i$ just gives $G_i\cong G/Z_i(G)$.
Indeed: you define $G_1=G/Z(G) = G/Z_1(G)$, so this holds.
Assume that $G_i\cong G/Z_i(G)$. You then define $G_{i+1}$ to be the group $G_i/Z(G_i)$. Now, $Z_{i+1}(G)$ is precisely the subgroup of $G$ such that $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$. Therefore, $Z_{i+1}(G)/Z_i(G)$ is precisely the group which corresponds to $Z(G_i)$ under the isomorphism $G/Z_i(G)\cong G_i$. Thus, $$G_{i+1} = \frac{G_i}{Z(G_i)} \cong \frac{G/Z_i(G)}{Z_{i+1}(G)/Z_i(G)} \cong \frac{G}{Z_{i+1}(G)}.$$ This establishes the claim.
So your question amounts to: if $G/Z_i(G)$ is abelian for some $i$, does it follow that $G$ is nilpotent? Answer: that's one of the standard definitions of "nilpotent group".