We want $BC=A$ such that $B$ and $C$ represent simple transformations (scaling etc).
As noted, all the points end up on the line $y=\frac 12 x$. Therefore either $B$ or $C$ must be a projection.
Using $A$ we see that point $(1,0)$ transforms to point $(2,1)$, and $(0,1)$ to $(4,2)$.
If we sketch these four points on a graph we see that the only realistic option for the other transformation is scaling. Moreover, the geometry suggests that scaling must happen first. Therefore $B$ is projection and $C$ is scaling.
Given that projection onto $y=\frac 12 x$ is orthogonal, point $(2,1)$ can be projected back to the $x$-axis (since it originated from $(1,0)$) perpendicular to $y=\frac 12 x$. And by looking at similar triangles, the $x$ scale factor can be found to be $\frac 52$.
The $y$ scale factor can be found similarly by projecting $(4,2)$ back to the $y$-axis.
From this $C$ can be assembled and then its inverse used to solve the matrix equation for $B$ (the projection).