There is exersice in Folland's book which show that every family $(A_{n})_{n=0}^{\infty}$in $\sigma$-Algebra,there exists a disjoint famly $(B_{n})_{n=0}^{\infty}$ such that $U_{n=0}^{\infty}A_{n}=U_{n=0}^{\infty}B_{n}$.
What is diffrence between the exersice and Vitali covering thereom?
you can find Vitali covering definition in the following link
The difference is clear if you look at the two statements. The hypotheses are totally different and the conclusions are totally different.
In the exercise the $A_n$ are elements of some $\sigma$-algebra, and the $B_n$ are not a subfamily of the $A_n$, although you don't say so, the $B_n$ are just required to be in the same $\sigma$-algebra.
In the "Vitali" theorem that you link to, which btw is not Vitali at all, it's really Wiener's lemma, a simpler substitute for Vitali, the $A_n$ are balls, the $B_n$ are a subfamily of the $A_n$, and the conclusion is not $\bigcup B_n=\bigcup A_n$, the conclusion is just $\bigcup A_n\subset \bigcup 3B_n$.
So asking what the difference is makes no sense - it seemms like you haven't read the two statements. They're simply not the same at all.
Why do we need Wiener's lemma if we can find $B_n$ with the same union? When we apply Wiener's lemma we need a subfamily of the $A_n$, and we need the $B_n$ to be balls - the exercise doesn't give either of those.