In one variable we have that the function approaches a line
$$f(x) = f(x_0) +f'(x_0)(x-a)$$
but in two variables we see that the function approaches a plane described by the two partial derivates:
$$z = z_0 + F_x(x-x_0) +F_y(y-yo)$$
First, how is this formula derived and second how is that $<F_x, F_y,1>$ is normal to the plane?
let:
$$L_1 = z_0+F_x(x-x_0)$$ $$L_2 =z_0 + F_y(y-y_0) $$
how would these two lines describe a plane? or how would you derive it by finding the normal vector of the plane
Excluding extreme circumstances such as the plane being perpendicular to the $xOy$ plane, a plane can be uniquely determined by $ax+by+cz+d (c\ne 0)$, thus given a point $(x_0, y_0, z_0)$ that the plane passes, we can use the formula $z(x,y) = z_0 + A(x-x_0) + B(y-y_0)$ to determine the plane.
Now using the definition of tangent, $\dfrac{\partial z}{\partial x} = \dfrac{\partial f}{\partial x}, \dfrac{\partial z}{\partial y} = \dfrac{\partial f}{\partial y} \Rightarrow A = F_x, B=F_y$.
The normal vector of the plane is simply $(F_x, F_y, 1)$ given the formula above. BTW $L_1$ and $L_2$ are also planes, so it's quite hard to use these to deduce the normal vector.