This question is about divided difference operators.
How do I perform $\partial _2$ or $\partial_3$ on $x_1^2x_2$?
$\partial_i$ is defined as $\frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,\dots,x_i,x_{i+1},\dots,x_n)=p(x_1,x_2,\dots,x_{i+1},x_i,\dots,x_n)$.
Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?
Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{\pi}=\partial_{\pi^{-1}w_0^n}S_{w_0^n}$. What does $\partial_{\pi^{-1}w_0^n}$ mean? Do I just concatenate $\pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $\pi=(12)$, we get $\partial_{\pi^{-1}w_0^n}=\partial_{1321}$?
For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence $$\partial_2(x_1^2x_2) = \frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $\partial_w$ you need to find a reduced word $I = [i_1, i_2, \ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} \dotsb r_{i_m}$ with $m$ minimal. Then $\partial_w = \partial_{i_1}\partial_{i_2} \dotsb \partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $\pi = 213$, hence $\pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $\partial_{\pi^{-1}w_0^3} = \partial_2\partial_1$.