Physical Interpretation of Relationship Between Hall Conductivity and Berry Curvature?

Why is the Hall conductivity in a 2D material $$\tag{1} \sigma_{xy}=\frac{e^2}{2\pi h} \int dk_x dk_y F_{xy}(k)$$ where the integral is taken over the Brillouin Zone and $F_{xy}(k)$ is the Berry curvature of the filled bands? What is the physical interpretation of this equation?

Also, can we re-parametrize all of the filled states by another pair of variables $A$ and $B$ and conclude that $$\tag{2} \sigma_{xy}=\frac{e^2}{2\pi h} \int F(A,B)dAdB$$ where $F(A,B)$ is the Berry curvature with respect to the $A$ and $B$ parameter space?

The formula follows from the Kubo formula of conductivity (based on the linear response theory), which is discussed in this question: Kubo Formula for Quantum Hall Effect and in the references therein. Starting from the Kubo formula (set $e=\hbar=1$) $$\tag{1}\sigma_{xy}=i\sum_{E_m<0<E_n}\frac{\langle m|v_x|n\rangle\langle n|v_y|m\rangle-\langle m|v_y|n\rangle\langle n|v_x|m\rangle}{(E_m-E_n)^2},$$ where $|m\rangle$ is the single particle eigen state of the eigen energy $E_m$, i.e. $$\tag{2} H|m\rangle = E_m|m\rangle.$$ Let us take the momentum derivative $\partial_\boldsymbol{k}$ on both sides of Eq. (2), we have $$\tag{3}(\partial_{\boldsymbol{k}}H)|m\rangle + H\partial_{\boldsymbol{k}}|m\rangle = (\partial_{\boldsymbol{k}}E_m)|m\rangle + E_m \partial_{\boldsymbol{k}}|m\rangle.$$ Then overlap with $\langle n|$ from left, Eq. (3) becomes $$\tag{4}\langle n|(\partial_{\boldsymbol{k}}H)|m\rangle + E_n\langle n|\partial_{\boldsymbol{k}}|m\rangle = (\partial_{\boldsymbol{k}}E_m)\langle n|m\rangle + E_m \langle n|\partial_{\boldsymbol{k}}|m\rangle.$$ Here we have used $\langle n|H = E_n\langle n|$. If $|m\rangle$ and $|n\rangle$ are different eigen states (for $E_m\neq E_n$ in Eq. (1)), their overlap should vanish, i.e. $\langle n|m\rangle=0$. Also note that $\partial_{\boldsymbol{k}} H$ is nothing but the velocity operator $\boldsymbol{v}=\partial_{\boldsymbol{k}} H$ by definition. So Eq. (4) can be reduced to $$\tag{5} \langle n|\boldsymbol{v}|m\rangle = (E_m - E_n) \langle n|\partial_{\boldsymbol{k}}|m\rangle.$$ Substitute Eq. (5) to Eq. (1) (restoring the $x$, $y$ subscript), we have $$\tag{6} \sigma_{xy}=-i\sum_{E_m<0<E_n}\big(\langle m|\partial_{k_x}|n\rangle\langle n|\partial_{k_y}|m\rangle - \langle m|\partial_{k_y}|n\rangle\langle n|\partial_{k_x}|m\rangle\big).$$
On the other hand, the Berry connection is defined as $\boldsymbol{A}=i\langle m|\partial_{\boldsymbol{k}}|m\rangle$, and the Berry curvature is $F_{xy}=(\partial_{\boldsymbol{k}}\times\boldsymbol{A})_z=\partial_{k_x}A_{y}-\partial_{k_y}A_{x}$. Given that $(\partial_\boldsymbol{k}\langle m|)|n\rangle = - \langle m|\partial_\boldsymbol{k}|n\rangle$ (integration by part), we can see $$\tag{7} F_{xy}= -i \sum_n\big( \langle m|\partial_{k_x}|n\rangle\langle n|\partial_{k_y}|m\rangle - \langle m|\partial_{k_y}|n\rangle\langle n|\partial_{k_x}|m\rangle\big) +i \langle m|\partial_{k_x}\partial_{k_y}-\partial_{k_y}\partial_{k_x}|m\rangle.$$ The last term will vanish as the partial derivatives commute with each other. So, by comparing with Eq. (6), we end up with $$\tag{8} \sigma_{xy}=\sum_{E_m<0}F_{xy}\sim\int_\text{BZ} d^2k F_{xy}.$$ This means that the Hall conductance is simply the sum of the Chern numbers, i.e. the total Berry flux through the Brillouin zone (BZ), for all the occupied bands. Of course, we are free to re-parameterize the momentum space by another pair of variables and the total Berry flux through the BZ will not change (as it is coordinate independent).
So what is the physical meaning of $F_{xy}$? $F_{xy}$ is an effective magnetic field in the momentum space (perpendicular to the $xy$-plane along the $z$-direction). We know that for the magnetic field $\boldsymbol{B}$ in the real space, a charged particle moving in it will experience the Lorentz force, such that the equation of motion reads $\dot{\boldsymbol{k}}=\dot{\boldsymbol{r}}\times \boldsymbol{B}$. Now switching to the momentum space, we just need to interchange the momentum $\boldsymbol{k}$ and the coordinate $\boldsymbol{r}$, and replace $\boldsymbol{B}$ by $\boldsymbol{F}$ (note that the symbol $\boldsymbol{F}$ here denotes the Berry curvature, not the force), which leads to $$\tag{9} \dot{\boldsymbol{r}}=\dot{\boldsymbol{k}}\times \boldsymbol{F}$$ So what is $\dot{\boldsymbol{r}}$? It is the velocity of the electron, which is proportional to the electric current $\boldsymbol{j}$. And what is $\dot{\boldsymbol{k}}$? It is the force acting on the electron (because the force is the rate that the momentum changes with time), which is proportional to the electric field strength $\boldsymbol{E}$, so Eq. (9) implies $$\tag{10} \boldsymbol{j} \sim \boldsymbol{E}\times \boldsymbol{F}.$$ Therefore the Berry curvature $F_{xy}$ at each momentum point simply gives the Hall response of the single-particle state at that momentum. So the Hall conductivity of the whole electron system should be the sum of the Berry curvature over all occupied states, which is stated in Eq. (8).