Every Boolean algebra $A$ is isomorphic to a field of set. In particular, if $A$ is finite, then $A$ is isomorphic to the power set of its atoms.
Now, suppose that $A$ is free Boolean algebra with 2 free generators (or atoms). Because $A$ is a finite Boolean algebra, it is isomorphic to the power set of its atoms. However, $A$ has $16$ elements, whereas the power set of its atoms has only 4 elements.
Could anybody explain what I seem to miss.
"Free generators" and "atoms" are not the same thing. An atom in a Boolean algebra is a minimal nonzero element, while a set of free generators is a set such that any map from the set to another Boolean algebra extends uniquely to a homomorphism of Boolean algebras. There is no reason to expect these to coincide.
In particular, if $B$ is freely generated by $\{x,y\}$, it turns out that $B$ has four atoms, namely $x\wedge y$, $\neg x\wedge y$, $x\wedge \neg y$, and $\neg x\wedge\neg y$. More generally, the free Boolean algebra on a finite set with $n$ elements has $2^n$ atoms, namely all the ways to form meets of the generators or their complements.