We are given basis B = (u1, u2) and basis C = (v1,v2).
u1 = (-1,2), u2=(2,-1),
v1=(1,0), v2=(1,1)
The coordinate vector x with respect to b is [x]B = (1,3) and coordinate vector x with respect to c is [x]C = (6, -1)
The example in the textbook asks: Using the bases B and C, find [x]C given that [x]B = (1,3)
This is where I get lost:
u1 = (-1, 2) = -3(1,0) + 2(1,1) = -3v1 +2v2
u2 = (2,-1) = 3(1,0) - (1,1) = 3v1 - v2
Where are they getting the coefficients -3 and 2 in u1?
Where are they getting the coefficients 3 and -1 in u2?
They don't explain it anywhere.
In this case they probably just got them out of their head since it's not that difficult to see.
But if you want a methodic approach, just set $$ (-1,2)=\alpha(1,0)+\beta(1,1)\tag{1} $$ with $\alpha,\beta$ to determine. Now $(1)$ is equivalent to $$ -1=\alpha+\beta\\ 2=\beta $$ Well, you should be able to solve this linear system!
Same thing for $u_2$ obviously.