Computing an image

Generators. Since $(x,y)=x(1,0)+y(0,1)$, we have $f(x,y)=xf(1,0)+yf(0,1)$, so $\operatorname{Im} f=\operatorname{span}\big(f(1,0),f(0,1)\big)$ with
$$f(1,0)=(1,1,2), \qquad f(0,1)=(1,-1,0).$$

Independence. These two vectors are not proportional, so they are linearly independent.

Conclusion. $\operatorname{Im} f=\operatorname{span}\big((1,1,2),(1,-1,0)\big)$, a plane in $\mathbb{R}^3$, of dimension $2$.

Remark (equation of the plane). For every $(x,y)$, the image $(X,Y,Z)=f(x,y)$ satisfies $X+Y-Z=(x+y)+(x-y)-2x=0$. The plane $\{X+Y-Z=0\}$ has dimension $2$ and contains $\operatorname{Im} f$, which also has dimension $2$, so $\operatorname{Im} f=\{(X,Y,Z): X+Y-Z=0\}$.