I fail to understand why we assume that $$\xi^{2}=A, \quad \xi^{2}=B \quad \Longrightarrow \quad A=\xi^{2}, \quad B=\xi^{2}$$ in the process of obtaining a solution to a PDE.
Example 4 Find the solution of the Cauchy problem governed by the linear PDE $$ \frac{\partial u}{\partial x}+2 x \frac{\partial u}{\partial y}=2 x u $$ subject to the conditions (a) $u(x, 0)=x^{2}$ for all $x$ and (b) $u(0, y)=y^{2}$ for all $y$. Solution The given PDE is of the form $$ a \frac{\partial u}{\partial x}+b \frac{\partial u}{\partial y}=c $$ where $$ a=1, \quad b=2 x, \quad c=2 x u $$ From the Lagrange-Charpit equations (8), we have $$ \frac{d x}{1}=\frac{d y}{2 x}=\frac{d u}{2 x u} $$ (a) Using the Lagrange-Charpit equation, we have the reciprocal of the slope of characteristic curves $$ \frac{d x}{d y}=\frac{1}{2 x} $$
Separating the variables and integrating to obtain $$ x^{2}=y+A $$ where $A$ is an arbitrary constant. Further, we have $$ \frac{d u}{d y}=u $$ which leads to $$ u=B e^{y} $$ where $B$ is an arbitrary constant. Thus, $$ x^{2}-y=A, \quad u e^{-y}=B $$ is a two-parameter family of characteristic curves. For solution curves to pass through the initial data, $F(x)=u(x, 0)=x^{2}$, we set $$ \xi^{2}=A, \quad \xi^{2}=B \quad \Longrightarrow \quad A=\xi^{2}, \quad B=\xi^{2} $$ where $\xi$ is a constant ( $x$-intercept, in this case) that identifies a characteristic curve. Thus, the characteristic and solution curves through this part of the initial curve are $$ x^{2}=y+\xi^{2}, \quad u=\xi^{2} e^{y} $$ Eliminating $\xi$ from the second equation using the first yields $$ u(x, y)=\left(x^{2}-y\right) e^{y} $$
TO BE Continued below
Source: https://www.iist.ac.in/sites/default/files/people/IN08026/MoC_0.pdf
Since the given condition is $u(x,0)=x^2$, the characteristic passes through the point $(x,y,u)=(\xi,0,\xi^2)$. So you put those values into the formulas $x^2-y=A$ and $ue^{-y}=B$ to obtain $\xi^2-0=A$ and $\xi^2 e^{-0}=B$.