Partial Derivatives of $F(x,y,z)$ where $z = f(x,y)$ When finding the tangent plane of a surface, given by $z = f(x,y)$, one method for doing so is writing $z$ implicitly: $F(x,y,z) = z - f(x,y) = 0.$
It is well established that the tangent plane of $F(x,y,z) = 0$ at $P(a,b,c)$, in general, is 
(1) $$F_x(a,b,c)(x-a) + F_y(a,b,c)(y-b) + F_z(a,b,c)(z-c) = 0.$$
However, the corollary of this theorem is that if $z = f(x,y)$, then the tangent plane is given by 
(2) $$-f_x(a,b)(x-a) + -f_y(a,b)(y-b) + (z-c) = 0.$$

What I do not understand is why this follows from the (1). 

In particular, let's examine $F_x(a,b,c). F_x(a,b,c) = F_x(z - f(x,y))$. I am confused why $F_x(z) = 0$. I thought $z = f(x,y)$, so shouldn't $F_x(z) = F_x(f) = f_x(a,b)$? Why can we treat $z$ as a constant here if it is a function of $x$ (and $y$).   
 A: You are making it too complicated. With $F(x,y,z):=f(x,y)-z$,
$$\begin{align}
F_x(x,y,z)&=f_x(x,y),
\\F_y(x,y,z)&=f_y(x,y),
\\F_z(x,y,z)&=-1.\end{align}$$
Partial derivatives assume that only one variable is active, the others are seen as constants.

Addendum:
If you consider that the functional relation between $z$ and $f(x,y)$ holds while taking the partials, you arrive at an absurdity: if $z=f(x,y)$,
$$F(x,y,z)=0$$
and
$$F_x(x,y,z)=F_y(x,y,z)=F_z(x,y,z)=0.$$
A: $F$ is a function on $\mathbb{R}^3$. Then level surface $F^{-1}(0) =\{
(x,y,f(x,y))\}$ has a normal vector : $$ {\rm grad}\ F = (-
f_x,-f_y, 1) $$
Hence at a point $P(a,b,c)$, $$ {\rm grad}\ F(P)\cdot
\bigg((x,y,z)-P\bigg) =0 $$
A: You say you are "confused why $F_x(z) = 0$", then conclude with "shouldn't $F_x(z) = F_x(f) = f_x(a,b)$?" But in fact, in going from your equation (1) to equation (2), the exact replacement $F_x(a,b,c) = f_x(a,b)$ (actually, its negative, due to the particular definition of $F$) is made that you seem to be suggesting.  This I find confusing. 
You go on to ask "Why can we treat $z$ as a constant here if it is a function of $x$ (and $y$)?"  But $z$ is not a function of $x$ and $y$. The function depending on $x$ and $y$ is $\,f$ (which we are not given here in terms of $x$ and $y$), while $z = f(x,y)$ is simply its value at $(x,y)$. Along the same lines, the notation $F_x$ usually stands for the function $\frac{\partial F}{\partial x}$, so that $F_x(a,b,c) = \frac{\partial F}{\partial x}(a,b,c)$. In that sense, your conjectures involving $F_x(z)$ and $F_x(f)$ make no sense, since the domain of $F$ and its partial derivatives is some subset of $\mathbb{R}^3$.
There are two functions involved here, and $z$ is not one of them. You have $f:\mathbb{R}^2\to\mathbb{R}$ with $f(x,y) = r$, some real number, and $F:\mathbb{R}^3\to\mathbb{R}$ with $F(x,y,z) = z-f(x,y)$, subject to the level surface condition $F(x,y,z) = 0$. The function $f$ has two partial derivatives, $f_x$ and $f_y$, while $F$ has three partial derivatives, related to those of $f$ by $F_x = -f_x$, $F_y = -f_y$, and $F_z = 1$. Your equation (2) follows from equation (1) by making these replacements (or rather, their negatives) at the point $P = (a,b,c)$ where $(a,b)$ is in the domain of $f$, and $(a,b,c)$ is in the domain of $F$.
I hope this is helpful to you.
ADDENDUM 
It is clear by now that my answer has not been helpful, and in fact seems to have generated more confusion, as witnessed by the comments. I should have been more clear from the outset, I suppose, that I don't understand how to interpret your notations "$F_x(z)$" and "$F_x(f)$". 
You understand that $z = f(x,y)$ defines a surface, where $z$ is a dependent variable which varies with $x$ and $y$ via the function $f$. This is also the level surface $F(x,y,z) = 0$ of the function $F$ of three independent variables $x$, $y$, and $z$ defined by $F(x,y,z) = z - f(x,y)$. You seem to want to restrict the independent variable $z$ in the definition of $F$ to satisfy $z = f(x,y)$, which is true only on the level surface $F(x,y,z) = 0$, i.e., $F(x,y,f(x,y)) = f(x,y)-f(x,y) = 0$. But in determining the equation of the tangent plane to $F$ at one of its points, say $(a,b,c)$, we do not restrict $F$ to a particular level surface in taking its gradient. The independent variables $x$, $y$, and $z$ are allowed to vary freely over the domain of $F$, which yield the usual partial derivatives and results mentioned in the last paragraph of my earlier answer (and in other answers to the question).
