Understanding how vector normal to tangent plane is of form $$, also about Stokes theorem My book says that a vector normal to a tangent plane at point $(a,b)$ is $$<f_x(a,b), f_y(a,b), -1>$$
If I understand vectors correctly, wouldn't this mean that the vector always moves $1$ unit down in the $z$ coordinates? But this should be obviously untrue, as the normal vector to a point in a tangent plane can be going in any direction, depending on the plane. And my books gives an image where it seems to do just that as an example! 

Obviously over here the normal vector is not moving $1$ unit down in the z-coordinates. So what does it mean that a vector normal to a tangent plane at point $(a,b)$ is of the form $<f_x(a,b), f_y(a,b), -1>$?
Also, my book gives the derivation of Stokes Theorem and says that the normal vector to the surface is $<-f_x(x,y), -f_y(x,y), 1>$. Why are the signs switched? Also , when my book discusses the flux of a vector field, it does the same thing and says "For the normal vector to point upward, we need a
positive z-component." But in the image I pasted, it certainly looks like the normal vector is pointing upward, and yet it is of the form $<f_x(x,y), f_y(x,y), -1>$
 A: There is not just one normal vector.  Any non-zero multiple of a normal vector is another normal vector, and that includes multiplying by $-1$.  
A: The issue here is $orientation.$ The normal vector shown in your graph is the one (presumably of unit norm) that points $away$ from the surface. You need to choose an orientation before doing any calculations, so that you can keep track of the "pieces" of the surface that you will want to integrate over. 
So, once and for all, we will always choose the vector of unit norm that points away from the surface.
Maybe the best way to see this is to look at an example. Take $f(x,y)=4-x^2-y^2$ at $(1,1,4)$. These data fit your graph fairly well.
Now, $f_x(1,1)=f_y(1,1)=-2$ so one possible normal vector is $\vec n=(-2,-2,-1)$. However, this choice of $\vec n$ points $into$ the surface, which violates our original choice of orientation. So, we simply switch its sign, multilpying by $-1$, to obtain $\vec n_1=(2,2,1)$, which, after normalization, is the vector shown in your graph. 
An interesting question is: is such a choice of orientation always possible? For example, consider the Mobius band and a point $P$ on it, attach a normal vector at $P$. How would you define the normal so that it points "away" from the surface?
