Consider the following fragment from Murphy's book "$C^*$-algebras and operator theory":
How can we extend $w_0$ to a map $w_1: H \to H$? I guess we first extend $w_0$ to the closure $\overline{v^*(H)}$ and then define it to be $0$ on its orthogonal complement. Does that work?
Also, why do we have $u = vw_1$? Does this depend on an explicit extension of $w_0$ or is it true for all bounded extensions of $w_0$?
Yes, you can always do that. If you have never done it, you need to show that any bounded linear operator can be extended by continuity to the closure of its domain. And that if you decompose $H=H_1\oplus H_1^\perp$ and $S,T$ are bounded operators on $H_1$ and $H_1^\perp$ respectively, then $S\oplus T$ defined by $(S\oplus T)(x\oplus y)=Sx\oplus Ty$ is a bounded operator with $\|S\oplus T\|\leq\|S\|+\|T\|$.