The most basic example of central extension with concrete groups and elements if possible I have read that the Heisenberg group is the central extension of the additive group as can be confirmed in this Wikipedia article This question is not about the Heisenberg group. It is about the central extensions of a group.  I wish I could sit here and provide more context with examples of something meaningful. The truth is that this concept is above me at this moment and I just need a gentle explanation with an example. I will ask questions in the comments if I am lost. The end goal of understanding this will be to take it back to the arena of the Heisenberg group and try to understand what is going on there.
If there is something wrong with the question, please let me know and I can try to fix it immediately as comments come in.
 A: One example which demonstrates the general theory quite well is the quaternion group $Q$ of $8$ elements. For concreteness, our elements are $\pm1,\pm i,\pm j, \pm k$, and multiplication is given by $i^2=j^2=k^2=-1$, and $ijk=-1$.
The centre of this group is $\pm 1$, and the quotient is the Klein four group, with $\bar i \cdot \bar j =\bar k$, and all square to the identity in this quotient group. One can think about it as “The Klein four group with some funny signs introduced in the multiplication”.
This is in general how central extensions can be interpreted, you have your quotient group $G$, and an abelian group $A$, which for simplicity, let’s take to be cyclic, consisting of some group of roots of unity.
Then a central extension of $G$ by $A$ is a group $\tilde G$ which behaves like $G$ with some funny roots of unity “twisting” the multiplication. More precisely, we can find a bijection between $\tilde G$ and $A\times G$ such that $A\times e$ behaves like “scalars” under multiplication, which is to say, acts centrally, and when we project onto the second coordinate, we just get multiplication in our group $G$. For example, in the quaternions, our $A$ is $\pm 1$, and under the bijection sending $(1, \bar i)$ to $i$, $(1, \bar j)$ to $j$, and $(1,\bar k)$ to $k$, our multiplication in $Q$ is: $$(1,\bar j) \cdot (1,\bar i)=(-1,\bar k)$$
Showing we have this funny twist by scalars thing happening. When one makes this notion precise (what is a twisting?? When do they give isomorphic groups $\tilde G$??) one is lead to the classification of central extensions by group cohomology, but without worrying about that, thinking of a central extension as a “scalar twisting” of the group multiplication law is very helpful.
