Please refer to the image given below (Book: Singular Points of Plane Curves by C.T.C. Wall).
Example 3.3.1:
For the third blow up, how do we come to the conclusions that $E_{0}$ is $x_{3}=0$ and the transform of $E_{1}$ does not meet the domain of this chart?
Working for the final blow up.
For the last blow up, I have two options (1) $(x_{5},y_{5})=(x_{6},x_{6}y_{6})$ or (2) $(x_{5},y_{5})=(x_{6}y_{6},y_{6})$.
Option (1) gives the exceptional curve $E_{5}: x_{6}=0$, transform of $E_{3}$ is not in this chart, transform of $E_{4}$ is $y_{6}=0$ and the transform of the curve is $C^{(6)}: y_{6}=1$.
Option (2) gives the exceptional curve $E_{5}: y_{6}=0$, transform of $E_{3}$ is $x_{6}=0$, transform of $E_{4}$ is not in this chart and the transform of the curve is $C^{(6)}: x_{6}=1$.
By definition, the blow-up of the point $(0,0)\in \mathbb{A}^2$ is given by
$$\begin{array}{ccc} \{([u:v],(x,y))\in \mathbb{P}^1\times \mathbb{A}^2\mid uy=vx\}&\to& \mathbb{A}^2\\ ([u:v],(x,y)) & \mapsto & (x,y)\end{array}$$
The exceptional divisor is then equal to $\mathbb{P}^1\times (0,0)$ and you have two charts on the blow-up: either $u\not=0$ or $v\not=0$. Choosing respectively $u=1$ and $v=1$ gives $y=vx$ and $x=uy$, so this gives two maps $$\begin{array}{ccc} \{([u:v],(x,y))\in \mathbb{P}^1\times \mathbb{A}^2\mid uy=vx\}&\to& \mathbb{A}^2\\ ([1:v],(x,vx)) & \mapsto & (x,vx)\\ ([u:1],(uy,y)) & \mapsto & (uy,y)\end{array}$$ and you can see then the blow-up in two charts: $(a,b)\mapsto (ab,b)$ or $(a,b)\mapsto (a,ab)$ from $\mathbb{A}^2$ to $\mathbb{A}^2$. In the first one, you remove the strict transform of $y=0$ and in the second one you remove the strict transform of $x=0$. Moreover, the exceptional curve is given by $b=0$ and $a=0$ respectively.
Looking at the text that you give, the author says that he has $(x_2,y_2)=(x_3y_3,y_3)$. Hence, the strict transform of $y_2=0$ is removed from the chart and the exceptional divisor is $y_3=0$.