When Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$? let there be $A,B \in V$.

$$A+B=\{a+b: a\in A,b \in B\}$$
Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$?
I think it is right, even if there are vectors that are both in $A$ and $B$.

Your notation is unclear. You probably mean $A$ and $B$ are subsets of $V$. If we denote by $A+B=\{a+b: a\in A, b\in B\}$, then a vector $v$ belongs to $\def\span{\operatorname{span}}\span(A+B)$ if it can be written as
$$
v=\alpha_1(a_1+b_1)+\alpha_2(a_2+b_2)+\dots+\alpha_n(a_n+b_n)
$$
for scalars $\alpha_1,\alpha_2,\dots,\alpha_n$ and $a_1,a_2,\dots,a_n\in A$ and $b_1,b_2,\dots,b_n\in B$. In this case
$$
v=(\alpha_1a_1+\alpha_2a_2+\dots+\alpha_na_n)+
(\alpha_1b_1+\alpha_2b_2+\dots+\alpha_nb_n)
$$
and this proves $\span(A+B)\subseteq\span(A)+\span(B)$.
If the vector space $V$ is finite dimensional, in order to have the reverse inclusion the condition is
\begin{align}
\dim\span(A+B)&=\dim(\span(A)+\span(B))\\
&=\dim\span(A)+\dim\span(B)-\dim(\span(A)\cap\span(B))
\end{align}
This is not true in general; an easy example is $A=\{u\}$, $B=\{-u\}$, where $u\ne0$. Then $A+B=\{0\}$, while $\span(A)=\span(B)=\langle u\rangle$ and, clearly,
$$
\{0\}=\span(A+B)\ne\span(A)+\span(B)=\langle u\rangle.
$$
