The normal to the paraboloid plays no role in this. A "surface element" ${\rm d}(x,y)$ at the point $(x,y)$ in the $(x,y)$-parameter plane produces via $\vec f$ (or rather $\vec f_*$) a surface element $dS$ at the point $\vec f(x,y)$ on your paraboloid $S$, and then this surface element $dS$ casts a shadow $d\omega$ on the unit sphere $S^2$ through central projection from $O$, i.e., via normalization of $\vec f$. Since $$|\vec f(x,y)|^2=x^2+y^2+{1\over4}(1-x^2-y^2)^2={1\over4}(1+x^2+y^2)^2$$ it follows that the shadow on $S^2$ is produced by the map $$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)\ .$$ This $\vec g$ is nothing else but an (unusual) parametric representation of $S^2$. In order to compute the area of the shadowed part of $S^2$ one has to compute $d\omega=|g_x\times g_y|{\rm d}(x,y)$ and to integrate this over the intended rectangle in the $(x,y)$-plane.
The computation gives, as already remarked by Ben, $$d\omega={4\over(1+x^2+y^2)^2}{\rm d}(x,y)\ .$$ Transforming to polar coordinates one finds for the $[-1,1]^2$-rectangle the exact value $8\sqrt 2\ \arctan(1/\sqrt 2)\doteq 6.96366$.
The normal to the paraboloid plays no role in this. A "surface element" $dA$ {\rm d}(x,y)$at the point$(x,y)$in the$(x,y)$-parameter plane produces via$\vec f$(or rather$\vec f_*$) a surface element$dS$at the point$\vec f(x,y)$on your paraboloid$S$, and then this surface element$dS$casts a shadow$d\omega$on the unit sphere$S^2$through central projection from$O$, i.e., via normalization of$\vec f$. In order to compute the size of Since $$|\vec f(x,y)|^2=x^2+y^2+{1\over4}(1-x^2-y^2)^2={1\over4}(1+x^2+y^2)^2$$ it follows that the shadow we have to consider on$S^2$is produced by the map $$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)$$ and bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)\ .$$This \vec g is nothing else but an (unusual) parametric representation of S^2. In order to compute its the area of the shadowed part of S^2 one has to compute d\omega=|g_x\times g_y|{\rm d}(x,y) .and to integrate this over the intended rectangle in the (x,y)-plane. 2 added 60 characters in body The normal to the paraboloid plays no role in this. A "surface element" dA at the point (x,y) in the (x,y)-parameter plane produces via \vec f a surface element dS at the point \vec f(x,y) on your paraboloid S, and then this surface element dS casts a shadow d\omega on the unit sphere S^2 via normalization of \vec f. So you In order to compute the size of that shadow we have to consider the map$$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)$$and to compute its$|g_x\times g_y|$d\omega=|g_x\times g_y|{\rm d}(x,y)$.