Here's how we get to the formula for the surface integral:
$$\Delta P_k=\frac{\Delta A_k}{\cos{\gamma}}$$
$$g:\text{ density }$$
$$\text{ Integral }=\sum_k \Delta P_k \cdot g(x_k, y_k, z_k) =\sum_k g(x_k, y_k, z_k) \cdot \frac{\Delta A_k}{\cos{\gamma}}=\iint_S g(x,y,z) \cdot \frac{dA}{\cos{\gamma}}$$
$$\cos{\gamma}=\frac{|\nabla{F} \cdot \hat{n}|}{|\nabla{F}|}$$
$$\iint_S g(x,y,z) \frac{|\nabla{F}|}{|\nabla{F} \cdot \hat{n}|} dA$$ $$d \sigma=\frac{|\nabla{F}|}{|\nabla{F} \cdot \hat{n}|} dA$$
So $$\iint_R g(x,y,z) d \sigma=\iint_S g(x,y,z) \frac{\nabla{F}}{|\nabla{F} \cdot \hat{n}|} dA$$
Could you explain me why at the beginning it is: $\Delta P_k=\frac{\Delta A_k}{\cos{\gamma}}$?
Could you also explain me the first and the last $"="$ of: $\text{ Integral }=\sum_k \Delta P_k \cdot g(x_k, y_k, z_k) =\sum_k g(x_k, y_k, z_k) \cdot \frac{\Delta A_k}{\cos{\gamma}}=\iint_S g(x,y,z) \cdot \frac{dA}{\cos{\gamma}}$?
At the second picture, why is it: $\hat{n}=\hat{k}$?
$(1)$
we are integrating on the region of projection of $P_k$ on $xy$ plane,
$(2)$ For the last line of $=$ we just change discrete summation into continuous integral by making $\Delta A_k$ infinitesimal.
$(3)$ $\hat n = \hat k$ because that is the direction of $xy$ plane and we are taking projection of the surface on $xy$ plane. so we take dot product of normal of surface to normal of $xy$ plane.
Added:: in $(2)$ me made a slight error. It's a dot product of vector function and the $P_k$ which has direction along $\frac{\nabla F}{|\nabla F|}$ i.e. unit normal vector.
See this image