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.
The $\Bbb C[x,y]$ case is a bit silly, because in that ring the only irreducible homogeneous polynomials are the ones of degree $1$.
The true version of the statement, as opposed to the current one which would be false with more variables because it allows $f_n$ and $f_{n-1}$ to have the same degree, is the following (henceforth, I'll indicate by $p_s$ the homogeneous component of degree $s$ of the polynomial $p$):
Let $R$ be an integral domain and let $f\in R[x_1,\cdots,x_r]$ be a polynomial such that $f_{\deg f}$ is irreducible. Then, $f$ is irreducible.
This is an obvious consequence of the identity $(pq)_{\deg(pq)}=p_{\deg p}q_{\deg q}$: on one hand, $f$ isn't a unit because $f_{\deg f}\notin R^*$ prevents it from being an invertible constant; on the other hand, if $f=pq$, then $f_{\deg f}=p_{\deg p}q_{\deg q}$ and therefore either $p_{\deg p}\in R^*$ or $q_{\deg q}\in R^*$. Either way, one of the factors $p,q$ is an invertible constant.
Best Answer
The Nullstellensatz precisely says that $f_1$ and $f_2$ have the same zero locus if and only if the ideals $(f_1)$ and $(f_2)$ have the same radical. You can easily see however (using unique factorization in $K[x_1,\cdots,x_n]$) that the radical of $(f)$ is the ideal generated by the product of the distinct irreducible factors of $(f)$. Hence $f_1$ and $f_2$ must have the same irreducible factors.