A linear projection

Linearity. Let $u=(x,y)$, $v=(x',y')$ and $\lambda,\mu\in\mathbb{R}$. Then
$$p(\lambda u+\mu v)=p(\lambda x+\mu x',\ \lambda y+\mu y')=(\lambda x+\mu x',\ 0)=\lambda(x,0)+\mu(x',0)=\lambda p(u)+\mu p(v).$$

Kernel. $p(x,y)=0\iff x=0$, so $\ker p=\{(0,y)\}=\operatorname{span}\big((0,1)\big)$, the $y$-axis.

Image. $\operatorname{Im} p=\{(x,0): x\in\mathbb{R}\}=\operatorname{span}\big((1,0)\big)$, the $x$-axis.

Idempotence. $p(p(x,y))=p(x,0)=(x,0)=p(x,y)$.

Conclusion. $p$ is the projection onto the $x$-axis parallel to the $y$-axis, and $\mathbb{R}^2=\ker p\oplus\operatorname{Im} p$.