Graph of a morphism of affine schemes Let $f:X\to Y$ be a morphism of schemes (here smooth, separated, of finite type over a field). Then the graph $\Gamma_f$ is defined as the image of the morphism $(id,f):X\to X\times Y$. 
Claim: The structure sheaf of the graph is a coherent $\mathcal O_{X\times Y}$-module. 
I tried to convince myself of this by considering the affine case first and got embarassingly confused:
So if $X=Spec\ (R)$ and $Y=Spec\ (S)$, then $X\times Y=Spec\ R\otimes S$. Let $f':S\to R$ be the corresponding ring homomorphism, then the graph should be just $\{s\otimes f'(s)\}$. And I guess I did something wrong here, because this is certainly no $S\otimes R$-module. The module generated by the set $\{s\otimes f'(s)\}$ on the other hand is the whole ring, because $f'(1)=1$.
It is easy to see that the graph is a closed subscheme, so it should be possible to write it as a quotient of $R\otimes S$. 

The graph $\Gamma_f\subset X\times Y$ is a closed subscheme of $X\times Y$ and as such corresponds to an ideal  of the ring $R\otimes_k S$, as you very correctly stated.
That ideal is the kernel $I$ of the ring morphism $$ R\otimes_k S\to R: r\otimes s\mapsto rf'(s) \quad (\star) $$The structure sheaf $\mathscr O_{\Gamma_f} $ of the graph is then the $\mathscr O_{X\times_kY}$- module $\widetilde {\frac {R\otimes_k S}{I}}$ corresponding to the $R\otimes_k S $- module $\frac {R\otimes_k S}{I}$  and is thus coherent.
 [Because $R\otimes_k S$ is noetherian. Beware that the structure sheaf of a scheme, even an affine one, needn't be coherent in the absence of noetherianness]
Note that the first projection $X\times_kY\to X$ restricts to an isomorphism $\Gamma_f \stackrel {\cong}{\to} X$ corresponding to the inverse of the isomorphism $\frac {R\otimes_k S}{I} \stackrel {\cong}{\to}  R $ deduced from $(\star) $.
