Question: The question asks to find a cubic equation given point A (0/18) slope here is 0. Point B only given the x=20 and slope is -0.6.

Please help, been trying to solve it for so long now. :(

1.) y=ax^3+bx^2+cx+d

2.) y'=3ax^2+2bx+c

-0,6=1200a+40b+1c

18=0a+0b+0c+1d

Answer: 
I think it's much easier to understand if you write the definition as "$Df\_{|a}$ is the unique linear map $L$ (if it exists) satisfying $f(x) = f(a) + L(x-a) + o(||x-a||)$ when $x\rightarrow a$".

Maybe it's even clearer if you write "$Df\_{|a}$ is the linear part of the unique affine map $A$ (if it exists) satisfying $f(x) = A(x) + o(||x-a||)$ when $x\rightarrow a$". 

So you can see that $A$ is the best possible affine approximation of $f$ near $a$ (because the error you make by replacing $f$ with $A$ is negligible compared to any affine map), and $Df\_{|a}$ is the linear part of this affine approximation.
The division by $\|h\|$ here is exactly analogous with the division by $h$ in the definition of the (standard) derivative of a real-valued function of a real variable:

$\dfrac{df}{dx}=\underset{h\rightarrow 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

To answer the second question, $Df|\_a$ is linear because it satisfies the linearity property, that is it commutes with addition and scalar multiplication. This is a consequence of how it is defined and is not actually the hard to prove (try using the definition to show $Df|\_a+Dg|\_a=D(f+g)|\_a$ and $D(\alpha f)|\_a=\alpha Df|\_a$ directly; hint, use some linear algebra)

To answer the final question, it is the natural analog of the familiar derivative of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for the case of a function from $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$.