The density function is symmetric about the $y$-axis, so by symmetry, the centre of mass has $x$-coordinate equal to $0$.
For the $y$ coordinate we (i) find the moment about the $x$-axis; (ii) find the mass; (iii) divide the moment by the mass.
You correctly found the mass.
For the moment about the $x$-axis, take a thin horizontal strip, from height $y$ to height $y+dy$. The strip has mass roughly $(2x)(1+y/2)\,dy$, and therefore moment roughly $(2x)(1+y/2)(y)\,dy$, where $x=\sqrt{4-y^2}$. Integrate from $y=0$ to $y=2$.
Or else, if one is familiar with the double integral, as the OP shows you are, a little $dx \times dy$ rectangle centered at $(x,y)$ has moment about the $x$-axis of about $(1+y/2)(y)\,dx\,dy$. Find the integral of this over the half-disk.
Remark: The height of the centre of mass is the same as the height of the centre of mass of the first quadrant part of our half-disk. I would probably find the area and moment of that, so as to have fewer minus signs around.