Let $X = \mathbb A^2$. Say we blow up the origin $(0,0)$, and then blow up the intersection of $(x=0)$ with the exceptional divisor. The resulting space is the blow-up of $\mathbb A^2$ along what subscheme?


I guess that the first blow-up is at $(x,y)$, while the second extracts the same divisor as the blow-up of $(x,y^2)$. So the two-point blow-up should be the blow-up at the product ideal $(x,y)(x,y^2) = (x^2,xy,y^3)$. Is this right?



Iterated blow-ups above a point
The question isn't quite well-defined: if we blow up $\mathbb A^2$ along $\mathfrak m = \langle x,y\rangle$ or along $\mathfrak m^{k}$ for $k>1$, we get the same space.


It's handy that the spaces you're asking about are all toric varieties. So one can compute polyhedrally. Instead of $X = \mathbb A^2$, consider the monoid $\{(p,q)\ :\ p,q\in\mathbb N\}$ (whose monoid algebra has spectrum $X$). To blow it up the first time, chop off the corner, leaving only the elements of $\mathfrak m$, or maybe of $\mathfrak m^k$ (i.e. removing an isosceles right triangle of size $k$).


But now you want to blow up again, cutting off one of the two new corners. I'll be a bit sketchy here and just say that to have a complete description, we want the resulting shape to have $2+1+1$ edges (not fewer) and only integer vertices. That suggests that the first cut should give $\mathfrak m^2$, leaving vertices $(2,0)$ and $(0,2)$, and the second should leave vertices $(2,0),(1,1),(0,3)$. Which exactly correspond to your suggested generators.


