The surface you integrate over in the divergence theorem has to be closed. The two surfaces you mention, the "hemiellipse" and the disk, are not individually closed, but together form a closed surface. Hence the integral over both is zero, by the divergence theorem, so the integral over the hemiellipse is the negative of the integral over the disk.
At the bottom of the cylinder, $z=0$. Since $F_z(x,y,0)=0$ we find
$$\begin{align}
\int_{S_1} \vec F \cdot \hat n\, dS&=\int_{S_1}\vec F(x,y,0)\cdot \hat z\,dS\\\\
&=0
\end{align}$$
and there is no flux contributed.
At the top of the cylinder, $z=5$. Since $F_z(x,y,5)=15$ we find
$$\begin{align}
\int_{S_2} \vec F \cdot \hat n\, dS&=\int_{S_2}\vec F(x,y,5)\cdot \hat z\,dS\\\\
&=\int_0^{2\pi}\int_0^2 (15)\,\rho\,d\rho\,d\phi\\\\
&=60\pi
\end{align}$$
Putting it all together, we find
$$\begin{align}
\int_S \vec F\cdot \hat n \,dS&=\int_V \nabla \cdot \vec F\,dV-\int_{S_2} \vec F \cdot \hat n\, dS-\int_{S_1} \vec F \cdot \hat n\, dS\\\\
&=120\pi-60\pi-0\\\\
&=60\pi
\end{align}$$
Best Answer
$$\iint_S\vec F\cdot \vec ndS=\iiint_V\nabla\cdot \vec FdV=3\iiint_V(x^2+y^2+z^2)dV$$ $$=3\iiint_Vr^2dV=3\int_0^14\pi r^4dr=\frac{12}{5}\pi$$