In Nielsen's book, the Kraus operator can be attained by trace out the enviroment: $$\operatorname{Tr}_{\rm env}[\hat{U}(|\psi\rangle\otimes|0\rangle)(\langle\psi|\otimes\langle 0|)\hat{U}^\dagger]. $$ And hence we can define the Kraus operator as $\Pi_l = \langle l|\hat{U}|0\rangle$.
But why can we write the total state $|\Psi\rangle=\hat{U}(|\psi\rangle\otimes| 0\rangle)$ as $|\Psi\rangle = \sum_l(\Pi_l|\psi\rangle)\otimes|l\rangle$?
You have $$\newcommand{\ket}[1]{\lvert #1\rangle}U(\ket\psi\otimes\ket0) = \bigg(I\otimes \underbrace{\sum_\ell \ket\ell\!\langle\ell|}_{\equiv I}\bigg) U (\ket\psi\otimes\ket0) \\ = \sum_\ell (I\otimes \ket\ell\!\langle\ell|)U(\ket\psi\otimes\ket0) = \sum_\ell (U_{(\ell,0)}\ket\psi)\otimes\ket\ell $$ where $$\Pi_\ell\equiv U_{(\ell,0)} \equiv (I\otimes \langle\ell|)U(I\otimes\ket0).$$ To be more explicit, decomposing in components the expression you get $$ (I\otimes \ket\ell\!\langle\ell|)U(\ket\psi\otimes\ket0) = \sum_{i,j} \ket{i,j} (\langle i|\otimes\langle j|)U(\ket\psi\otimes\ket0) \\ = \sum_{i,j} \ket{i,j}\, (\langle i|\otimes I) \underbrace{(I\otimes\langle j|) U (I\otimes \ket0)}_{\equiv U_{(j,0)}} (\ket\psi\otimes I) \\ = \bigg(\underbrace{\sum_i \ket i\!\langle i| U_{(j,0)}\ket\psi }_{=U_{(j,0)}\ket\psi}\bigg) \otimes \ket j $$
