How to use the Kraus operators to represent the total density matrix instead of the reduced one?

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 $$(I\otimes \ket\ell\!\langle\ell|)U(\ket\psi\otimes\ket0) = \sum_{i,j} \ket{i,j} (\langle i|\otimes\langle j|)U(\ket\psi\otimes\ket0) \\ = \sum_{i,j} \ket{i,j}\, (\langle i|\otimes I) \underbrace{(I\otimes\langle j|) U (I\otimes \ket0)}_{\equiv U_{(j,0)}} (\ket\psi\otimes I) \\ = \bigg(\underbrace{\sum_i \ket i\!\langle i| U_{(j,0)}\ket\psi }_{=U_{(j,0)}\ket\psi}\bigg) \otimes \ket j$$