I'm having a hard time grasping what they're saying in my textbook for this chapter.
If $z=f(x,y)$ defines a surface, then the tangent plane to the surface at (a,b) may be obtained as $$(-f_x(a,b),-f_y(a,b),1) * (x-a,y-b,z-f(a,b))=0$$ This is the result of applying the gradient-of-a-surface formula to the surface $z-f(x,y)=0$ at the point $(a,b,f(a,b))$.
How is the formula comparable to the formula for finding tangent planes to a surface which is $$(f_x(a,b,c),f_y(a,b,c),f_z(a,b,c))*(x-a,y-b,z-c)=0$$ and why are the $f_x$ and $f_y$ components negative in the first formula?
The two formulations are the same. We are taking $F(x,y,z)=z-f(x,y)$ and setting $F=0$. Taking partial derivatives of $F$ gives $$F_x(a,b,c)=-f_x(a,b)$$ $$F_y(a,b,c)=-f_y(a,b)$$ $$F_z(a,b,c)=1$$
The partial derivatives with respect to $x$ and $y$ have negative coefficients because we are taking the gradient of $z-f(x,y)$. They would have positive coefficients if we defined $F(x,y,z)=f(x,y)-z$ but then $F_z(a,b,c)$ would be $-1$ instead of $1$.