Let $F(x,y,z) = \langle x, y, z^2 \rangle$, and $S$ be the unit sphere with radius $1$ about the origin. The question is to find the surface integral $\iint_S F \cdot \text{d} S$. Using the divergence theorem we can compute it to be $8 \pi /3$, but suppose we split the sphere into upper and lower hemispheres and use the formula
$$\iint_S F \cdot \text{d} S = \iint_D \left( -F_1 \partial_x g - F_2 \partial_y g + F_3 \right)\text{d} A$$
where $F_j$ are the components of $F$, $D$ is the unit circle about $0$ in the $xy$-plane, and $g(x,y) = \pm \sqrt{1-x^2-y^2}$. Once we split and add the two integrals we get (the first two terms get cancelled, and $F_3=z^2$ gets multiplied by $2$, because it doesn't see the sign of the surface function $g$)
$$\iint_S F \cdot \text{d} S = 2 \iint_D (1-x^2-y^2) \text{d} A = 4\pi \int_0^1 (1-r^2) r \text{d} r = \pi$$
which is a wrong answer. Where am I making a mistake?
If you are integrating a divergenceless vector field over a region bounded by two surfaces, then by Stokes' theorem, the flux into one surface equals the flux out of the other. In that case, when computing the flux through a hemisphere, you could opt to replace it by the equatorial disk, as you have attempted to do.
But your vector field is not divergenceless. You have $\nabla \cdot F=2+2z$. The flux integrals of this field are not "independent of surface". You cannot replace the upper hemisphere by the equatorial disk.
Moreover, I'm not sure why you're integrating $\iint F\cdot\nabla g\,dA$. That is neither the flux through the hemisphere, nor through the equatorial plane.