The operator U defined as $U|\psi\rangle\otimes|0\rangle=\sum_m M_m|\psi\rangle\otimes|m\rangle$ where $\sum_mM_m^\dagger M_m=I$, preserves the scalar product $(\langle \phi|\otimes\langle 0|U^\dagger)(U|\psi\rangle\otimes|0\rangle)=\langle\phi|\psi\rangle=(\langle\phi|\otimes\langle 0|)(|\psi\rangle\otimes|0\rangle)$ where $|\psi\rangle,|\phi\rangle\in \mathcal{H}_A$ and $|0\rangle \in\mathcal{H}_B$ and $\{|m\rangle\}_{m=1}^m$ form an orthonormal basis for $\mathcal{H}_B$. $U$ is a unitary operator in the one dimensional subspace spanned by $|0\rangle$. Besides, it may be extended to a full unitary operator in the global Hilbert space $\mathcal{H}_A\otimes \mathcal{H}_B$.
Theorem 1: I understand the proof of the statement if $U:W\to V$ is a unitary operator, then this can be extended to the whole vector space such that there exists a unitary operator $U':V\to V$, where $W$ is a subspace in the vector space $V$. Check Extension of unitary operator in subsystem to complete system for the proof.
But, how do I make use of this in our context so that $U$ is defined for the entire statespace $\mathcal{H}_A\otimes \mathcal{H}_B$ ?
How do I make sense of the statement that since $U$ preserves the inner product as above, it is a unitary operator in the one dimensional space spanned by $|0\rangle$ ?
Original context from the Reference below (Page 12):
My Attempt
If $\mathcal{H}_A$ and $\mathcal{H}_B$ are vector spaces with basis $\{|\psi_i\rangle\}$ and $\{|m\rangle\}$ then $\mathcal{H}_A\otimes \mathcal{H}_B$ has a basis $\{|\psi_i\rangle\otimes |m\rangle\}$. Check "Can we recover the bases of two infinite-dimensional vector spaces into a tensor product?" for a proof of this.
$$ \langle\psi_i|\psi_j\rangle=\delta_{ij}\quad\&\quad \langle m|m'\rangle=\delta_{mm'}\\ \implies (\langle\psi_i|\otimes\langle m|). (|\psi_j\rangle\otimes|m'\rangle)=\langle\psi_i|\psi_j\rangle\otimes\langle m|m'\rangle $$
If $\{|\psi_i\rangle\}$ and $\{|m\rangle\}$ are orthonormal bases for $\mathcal{H}_A$ and $\mathcal{H}_B$, then $\{|\psi_i\rangle\otimes |m\rangle\}$ is an orthonormal basis for $\mathcal{H}_A\otimes \mathcal{H}_B$.
From this, we can conclude that $\{|\psi_i\rangle\otimes|0\rangle\}$ is a set of orthonormal vectors, thus form a subspace of $\mathcal{H}_A\otimes \mathcal{H}_B$.
Since $(\langle \psi_i|\otimes\langle 0|U^\dagger)(U|\psi_j\rangle\otimes|0\rangle)=(\langle\psi_i|\otimes\langle 0|)(|\psi_j\rangle\otimes|0\rangle)$ for all $|\psi_i\rangle$, $|\psi_j\rangle$, we can say $U$ is a unitary operator in the subspace spanned by $\{|\psi_i\rangle\otimes|0\rangle\}$. Using Theorem 1, this unitary operator can be extended to the whole vector space $\mathcal{H}_A\otimes \mathcal{H}_B$.