Homogenous polynomial and partial derivatives I'm struggling to understand this part in a book I'm reading:

Let $F$ be a projective curve of degree $d$ with $P\in F$. Wlog,
  suppose $P=(a:b:1)$. Let's look the affine chart $(a,b)\mapsto
 (a,b,1)$.
Let $f$ be the deshomogenization of $F$, we can write $f$ in this way:
  (WHY?)
$$f=F(x,y,1)=f_1(x-a,y-b)+\ldots +f_d(x-a,y-b)$$
Where $f_l$ is a homogeneous polynomial of degree $l$ and we have
  (WHY?) 
$$f_l=\sum_{i+j=l}\frac{1}{i!j!}\frac{\partial^lf}{\partial^ix\partial^jy}x^iy^j$$

I know this should be a silly question, but I'm a beginner in this subject and I really need help, if anyone could help me I would be grateful.
Thanks 
EDIT
I'm thinking about Taylor's formula, but The formula of the post doesn't match with the Taylor's  one of several variables, see for example this link, maybe there is some mistake in the formula of my post?

Question: "I know this should be a silly question, but I'm a beginner in this subject and I really need help, if anyone could help me I would be grateful. Thanks"
Answer: When trying to prove a formula you should verify the formula in an explicit "elementary " example first, then try to generalize.
Example: Let $F:=x^2+y^2+z^2$ and let $p:=(a,b,1) \in D(z)$ with $u:=x/z, v:=y/z$ local coordinates. Let $f(u,v):=u^2+v^1+1$.
It follows
$$f(u,v)=f(a+u-a, b+v-b)=(a+u-a)^2+(b+v-b)^2+1= $$
$$a^2+b^2+1+ 2a(u-a)+2b(v-b)+(u-a)^2+(v-b)^2=$$
$$=f(a,b)+\frac{\partial f}{\partial_u}(a,b)(u-a)+\frac{\partial f}{\partial v}(a,b)(v-b) + $$
$$\frac{\partial^2 f}{\partial u^2}(a,b)(u-a)^2 +\frac{\partial^2 f}{\partial u \partial v}(a,b)(u-a)(v-b)+\frac{\partial^2 f}{\partial v^2}(a,b)(v-b)^2.$$
In general you get for any polynomial $g(u,v)$ the formula
$$g(u,v)=T_p(g(u,v)):=\sum_{k\geq 0} \sum_{i+j=k}\frac{\partial^{k} f}{\partial^i_u \partial^j_v}(a,b)(u-a)^i(v-b)^j,$$
which is the Taylor expansion of $g(u,v)$ at the point $p$.
