Calculating the coordinate ring and irreducible components Consider the graded ring $S=(R/I)\oplus (I/I^2)\oplus (I^2/I^3)\oplus\cdots$
Take $R=k[X,Y],I=(X^2Y,XY^2)$.
Then $S=k[X,Y]/(X^2Y,XY^2)\oplus(X^2Y,XY^2)/(X^2Y,XY^2)^2\oplus\cdots$. 
I am not sure how to calculate the components of the scheme $\operatorname{Spec}(S)$ and the  multiplicities of each component?
I am not sure how to describe the ring, I guess it is $k[X,Y,S,T]/(X^2Y,XY^2,SXY^2,TXY^2,SX^2Y,TX^2Y,SY-XT)$? 
 A: $\newcommand{\ideal}[1]{\mathfrak{#1}}$
One can compute a representation of $S$ by considering the map
$$\phi:k[X,Y,U,V] \mapsto k[X,Y,T]$$
with $\phi(X) = X$, $\phi(Y)=Y$, $\phi(U)=T X^2 Y$, $\phi(V)=T X Y^2$.
It is $\operatorname{im}\phi = k[X,Y][T (X^2 Y) , T (X Y^2)]$ that is, the Rees-Algebra $B=A[t I]$ for $A=k[X,Y]$ and $I \subseteq A$, $I = (X^2 Y, X Y^2)$.
So the Rees-Algebra $B$ is equal to 
$$B' = k[X,Y,U,V]/(\ker \phi) = k[X,Y,U,V]/(X V - Y U) =: B''/J$$
Now $S$ (in the original notation) is $B/(IB)$ or $B'/IB'$ as $B \cong B'$ as $A$--Algebras. So we have
$$
\begin{multline}
S = B'/IB' = (B''/J)/(I (B''/J)) = B''/(J + I) = \\
= k[X,Y,U,V]/(X^2 Y, X Y^2, X V - Y U)
\end{multline}
$$
So your guess for $S$ is correct, you just put in some unnecessary extra generators.
If one specializes to $k=\mathbb{Q}$ and calculates with Macaulay2 a primary decomposition of the zero ideal in $S$ one gets
$$(0) = (X,U) \cap (Y,V) \cap (Y^3,X^3)$$
The prime-ideals belonging to these are $\ideal{p}_1=(X,U)$, $\ideal{p}_2 = (Y,V)$, $\ideal{p}_3 = (X,Y)$ in the same order. 
So the multiplicities of the first two components are 1 and the third is the length of $(S/(Y^3,X^3))_{\ideal{p}_3}$. This can not be easily calculated with Macaulay2, as it is in the localization ring $S_{\ideal{p}_3}$. Trying simply the degree of $(S/(Y^3,X^3))$ one gets $3$ for the degree, which is perhaps the multiplicity here (I am not quite sure if this is right).
So the exceptional divisor $E$ of the blowup of $k[X,Y]$ at $(X^2 Y, X Y^2)$ consists of three components, one over the $X$-axis, one over the $Y$--axis and one over $(X,Y)$, which is not unexpected considering the form of $I$.
