It is a theorem proven in Donald Monk's set theory notes (page 184) that if $S$ is a consistent set of sentences containing ZFC, then if we add a constant $\mathbf{M}$ to the language of set theory the following set of sentences will be consistent: $$ S\cup \{\mathbf{M} \textrm{ transitive}\} \cup \{|\mathbf{M}|=\omega\} \cup \{\phi^{\mathbf{M}}: \phi\in S\}. $$
Monk concludes from this theorem that there exist countable transitive models of ZFC. But how can this fact can be derived from the theorem?
My attempt to prove the existence of countable transitive models was to pick $S=\textrm{ZFC}$ and then use completeness to show the existence of a model $(\mathcal{M},R)$ such that $\mathcal{M} \vDash \textrm{ZFC}\cup \{\mathbf{M} \textrm{ transitive}\} \cup \{|\mathbf{M}|=\omega\} \cup \{\phi^{\mathbf{M}}: \phi\in \textrm{ZFC}\}.$ Then I was going to prove well-foundedness of the relation $R$, use Mostowski collapse to find a standard, transitive model $\mathcal{N}$ of the sentences, and then prove that $(\mathbf{M}^{\mathcal{N}},\in)$ had to be a countable, transitive model of ZFC. Unfortunately I learned from the responses to the post If $(M,R)\vDash \textrm{ZFC}$ must $R$ be a Well-Founded Relation? that we can't guarantee $R$ will be well-founded just from the fact that $\mathcal{M}$ models ZFC. So how can we justify Monk's assertion that his theorem proves the existence of countable transitive models of ZFC? Am I overthinking this?