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?
 A: User @adrido basically answered this question in the comments, but I thought I'd elaborate more fully, in case anyone encounters this in the future.
The polynomial $f$ consists of a a number of homogeneous parts - these are the collections of terms of $f$ which all have the same degree:
$$
\underbrace{x^3y^2 - x^5}_{\deg 5} + \underbrace{3x^2y^2 + 9y^4}_{\deg 4} + \underbrace{2xy^2 - 7x^3 + 15y^3}_{\deg 3} + \cdots
$$
The big question here is what on earth does this have to do with all those partial derivatives? You're right, it's not a Taylor series, but it's sorta the same idea.
By taking the $\ell$-th partial derivatives of a polynomial $f$ and evaluating it at $0$, what we're left with is the coefficients of the degree $\ell$ terms (the partial derivatives cause all $k<\ell$ degree terms to vanish, and all higher degree terms vanish when we evaluate at $0$), but the coefficients are multiplied by $i!j!$ where $i+j = \ell$. Indeed, you can check for yourself that for any polynomial $f(x, y)$, the coefficient of $x^iy^j$ in $f$ (which we will denote $[x^iy^j]f(x, y)$ is given by
$$
[x^iy^j]f(x, y) = \frac1{i!j!}\frac{\partial^{i+j}f}{\partial x^i\partial y^j}(0, 0)
$$
So we can reconstruct $f$ using these coefficients we've just obtained:
$$
f = \sum_{i,j}\left([x^iy^j]f\right)x^iy^j = \sum_{i,j}\left(\frac1{i!j!}\frac{\partial^{i+j}f}{\partial x^i\partial y^j}(0, 0)\right)x^iy^j
$$
I suppose what's nice about this decomposition is it collects up the homogeneous parts of $f$ as those terms above for which $i+j$ is constant, equal to some $\ell$.
Of course, everything above generalizes to a polynomial $f(x_1, ..., x_n)$, but I leave it to you to work out the (messy, but straightforward) details.
A: 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 $p$.
