Let $V$, $\dim V=n-1$ be the standard representation of the symmetric group $S_n$ and let $V'= \langle x_1,x_2,\ldots,x_n \rangle$ be its natural representation. Then ( see. Fulton, Harris, 4.19) we have $$ {\rm Sym}^2 V=U \oplus V \oplus V_{(n-2,2)}, $$ where $U$ is the trivial representation and $V_{(n-2,2)}$ is a representation that correspond to a partition $(n-2,2).$
Question 1. What is the decomposition $ {\rm Sym}^2 V'? $
Question 2. Can we indicate in explicit way a basis for each irreducible component of this decomposition?
My try is as follows. Since $V'=U \oplus V$ then $$ {\rm Sym}^2 V'={\rm Sym}^2U \oplus {\rm Sym}^2 V \oplus U \cdot V. $$ I think that ${\rm Sym}^2U \cong U $ and $U \cdot V \cong V.$ Then $$ {\rm Sym}^2 V'={\rm Sym}^2U \oplus {\rm Sym}^2 V \oplus U \cdot V=2 U \oplus 2 V \oplus V_{(n-2,2)}. $$ Am I right?
If yes, then what be answer for the second question? I know that $U=\langle x_1+x_2+\cdots+x_n \rangle$ and $V= \langle x_1-x_2,x_1-x_3,\ldots,x_1-x_n \rangle$, but I don't know what is the basis for its realisation in ${\rm Sym}^2V'$ as polynomials of degree $2.$
Edit.
$2U=\langle x_1^2+x_2^2+\cdots+x_n^2 \rangle \oplus \langle x_1 x_2+x_1 x_2 +\cdots+ x_{n-1} x_n\rangle $
Butwhat is the basis of $V_{(n-2,2)}$ and $V$ realised in ${\rm Sym}^2 V'?$