If [itex] x_1,...,x_n[/itex] is a basis of V, and [itex] \left< y, x_j \right> = 0[/itex] for all j, then y=0 necessarily.
Proof: Suppose that [itex] y = a_1 x_1 +... + a_n x_n [/itex] for some numbers a_{1},...,a_{n}. Then if y is nonzero, [itex] \left< y, y \right> \neq 0[/itex]. But
[tex] \left< y, y \right> = \left< y, \sum a_j x_j \right> = \sum a_j \left< y, x_j \right> = 0[/tex] since every inner product in that summation is zero. So y must have been zero to begin with
I'm not where your confusion is with y_{i} being in the orthogonal space. They haven't defined y_{i} to be anything in particular, they're just saying, suppose they happened to pick such a vector
It may help to do an example. Let's do it in 2 dimensions: x_{1} = (1,0) and x_{2} = (1,1). Then y_{1} is perpendicular to everything EXCEPT for x_{1}. In particular, y_{1} is perpendicular to x_{2} which means y_{1} must be of the form (a,a) for some number a. Then <x_{1},y_{1}> = a = 1 implies that a=1. So y_{1} = (1,1).
<y_{2},x_{1}> = 0 implies that y_{2} is of the form (0,c) for some number c. <y_{2},x_{2}> = c, so for the inner product to be 1 it must be c=1, So y_{2} = (0,1)
