First we show the direct sum.
To show that $V = U \bigoplus W$ we need to show that every element $v\in V$ can be written in the form $v = u + w$ where $u \in U$ and $w \in W$, and moreover we need to show that $U\cap W = 0$.
First let $v\in V$. Then we can write $v = (x,y)$ since $V \subset \mathbb{R}^2$ and hence $v = (x + y, 0) + (-y, y) \in W + V$ so we have showed the first part, and now we need to show that the intersection is the $0$ space.
Suppose $a\in U\cap V$ then write $a = (x,y)$; Since $a \in W$ that $y = 0$ so $a = (x,0)$ but then since $a\in U$ this implies that $x = -0 = 0$ and hence $a = (0,0) \in 0$ So we can now see that $V = U\bigoplus W$.
Then we need to define the projection onto $U$ and $W$, but these are defined exactly by the formula you cited in the fist line, and are well defined linear operators since $V = U \bigoplus W$. One can see this is well defined because, by the direct sum, we can uniquely write $v = u + w$ for all $v$, and hence we only have one possible value for the projection onto either $U$ or $W$ (namely $u$ or $w$ respectively)