This problem is two dimensional so dividing the x-axis into 2 parts and the y-axis into two parts divides the rectangle into 2(2)= 4 sub rectangles. Those have four "lower right corners" so you need four points, not two.
The rectangle $\displaystyle 0\le x\le 2$ and $\displaystyle 0\le y \le 4$ divides into 4 sub intervals, A: $\displaystyle 0\le x\le 1$, $\displaystyle 0\le y\le 2$; B: $\displaystyle 0\le x\le 1$, $\displaystyle 2\le y\le 4$; C: $\displaystyle 1\le x\le 2$, $\displaystyle 0\le y\le 2$; D: $\displaystyle 1\le x\le 2$, $\displaystyle 2\le y\le 4$. The lower right corner of each rectangle is A: (1, 0) where f(x,y)= 1(0)= 0; B: (1, 2) where f(x, y)= 2; C: (2, 0) where f(x,y)= 2(0)= 0; and D: (2, 2) where f(x,y)= 2(2)= 4. Further each rectangle has width 1, height 2 so area 2.
The Riemann sum is 2(0)+ 2(2)+ 2(0)+ 2(4)= 4+ 8= 12.