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.
This is just a matter of terminology. In both books I have to hand (Hartshorne, and Eisenbud's "Commutative algebra..."), the authors define an 'affine algebraic set' to be any subset of $\mathbb{A}^n$ given by the vanishing of polynomials, and an 'affine algebraic variety' to be an irreducible such set.
What is perfectly clear (and is possibly what 'The question' really asks, given the absence of the word 'variety' at the end) is that any closed subset of an affine algebraic set is again an affine algebraic set.
In practice though, the word 'variety' often seems to be used more generally, without the irreducibility requirement.
Best Answer
It is true that the dimension of $V$ as a vector space is equal to it's dimension as a variety. This is because $I(V)$ can be generated by $n-\dim V$ linear forms, where we mean dimension as a linear space.
To prove this, select a basis $v_1,\cdots,v_m$ for $V$ and complete it to a basis of $k^n$ with the vectors $w_{m+1},\cdots,w_n$. Writing these vectors as the columns of a matrix $M$, we see that this matrix is invertible and takes the standard basis vectors $e_i\mapsto v_i$ if $i\leq m$ and $e_i\mapsto w_i$ if $i>m$. So $M$ and $M^{-1}$ define an isomorphism between $V$ and the subspace $V'$ consisting of vectors with final $n-m$ coordinates zero (this is an isomorphism as vector spaces and also as varieties). This means that as the coordinate functions $x_{m+1},\cdots,x_n$ generate the ideal of $V'$, their pullbacks under $M^{-1}$ generate the ideal of $V$, and we see the claim.