Coordinate rings in projective spaces. What are they? When $X$ is an algebraic variety of affine $n$-space, then the coordinate ring of $X$ are polynomials restricted to $X$. 

But when $X$ is a variety in the projective $n$ space, what are the elements of its coordinate ring? 

I know that the coordinate ring for the projective space is given by $K[X_0,X_1,\dots,X_n]/I(X)$, where $I(X)$ is a homogeneous ideal. But what does elements in $K[X_0,X_1,\dots,X_n]/I(X)$ look like? Are they homogemeous polynomials in $K[X_0,X_1,\dots,X_n]$ restricted to $X$ or are they all polynomials $F$ in $K[X_0,X_1,\dots,X_n]$ restricted to $X$, with $F(la_0,la_1,\dots,la_n)=F(a_0,a_1,\dots,a_n)$ for all $l$ in algebrically closed field $K$ and $(a_0,a_1,\dots,a_n)$ belongs to $X$?
 A: The elements of $S=K[X_0,X_1,\dots,X_n]/I(X)$ are perfectly good functions...but not on $X$ !
They are functions on the cone $c(X)\subset \mathbb A^{n+1}$ associated to $X$, which has as equations the very elements of  $I(X)$.
For example if $X$ is the circle $X_0^2+X_1^2-X_2^2=0$ in $\mathbb P^2$, then $c(X)$ is the  cone $X_0^2+X_1^2-X_2^2=0$ in  $\mathbb A^3$.
Unfortunately the elements of $S$ are not constant on the generatrices of the cone, so that the functions in $S$ do not descend to $X$, just as Jyrki and Matt told you.
You can experiment with $\overline X_0 \in S$ in the example above: $\overline X_0 $ takes all  values in $K$ on the generatrix $K(1,0,1)\subset c(X)$, so that it does not make sense to attach a value to $\overline X_o$ at the point $[1:0:1]\in X\subset \mathbb P^2$.  
So the graded ring  $S$ is useless, right? Wrong!
It is a ring with which one can build  other interesting rings .
For example the field of rational functions $Rat(X)$ is the subfield of the fraction field $Frac(S)$ consisting of quotients $\overline P/\overline Q$  with $P,Q$ polynomials of the same degree and $ Q\notin I(X)$.
Similarly one can define the local ring $\mathcal O_{X,x}$ for $x\in X$ by adding the condition $Q(x)\neq 0$.
[You might look at this  answer  for the case $X=\mathbb P^n$ ]
By the way, note that even if you cannot attach a value to $Q(x)$, it makes perfectly good sense to say that $ Q(x)=0$ or $Q(x)\neq 0$.
And this is the source of more usefulness of the ring $S$: it permits you to define all subvarieties of $X$ by setting a bunch of homogeneous elements of $S$ equal to zero.  
To sum up, consideration of the graded ring $S$  allows you to cut loose from the ambient $\mathbb P^n$ and be more intrinsic.
And this is the beginning of the Proj construction, which is the generalization to scheme theory of the classical notion of projective variety.  
Edit:caveat
After all this praise of  $S$, I must draw the attention to one drawback : it does not depend only  on $X$ but also  on its embedding into $\mathbb P^n$:
If for example we embed $\mathbb P^1$ into $\mathbb P^2$ successively as the line $X_0=0$ and then as the circle $X_0^2+X_1^2-X_2^2=0$ already mentioned , the corresponding graded rings  are $S_1=K[X_1,X_2]$ and $S_2=K[X_0,X_1,X_2]/(X_0^2+X_1^2-X_2^2)$  and they  are not isomorphic.
A: The elements of the coordinate ring are sections of certain line bundles on $\mathbb P^n$ that are restricted to $X$. Unfortunately, there is not really a simpler answer than that, because they are not functions on $X$, just as homo. polys. are not functions on $\mathbb P^n$. 
