I (think) I've found the Heisenberg Lie algebra representation through quantization. Where we have $q \mapsto q$ and $p \mapsto -i \hbar \frac{\partial}{\partial q}$.
So this is only a Lie algebra representation. $\rho_*: \mathfrak{h} \to \text{End}(V) $.
Where $V$ is the hilbert space (representation space). And $\mathfrak{h}$ is the Heisenberg Lie algebra.
How do I lift this to a lie group representation $\rho: H \to \text{Gl}(V)$ . Where $H$ is the Heisenberg group? And how do I show it's unitary?
You are representing the three algebra elements $q, p, 1\!\!1$ by hermitian operators acting on functions of q. For simplicity, non-dimensionalize $\hbar=1$--only the clueless keep it unscaled.
Multiplying linear combinations of such operators by i and exponentiating yields the generic unitary group element representation for you , $$ e^{ic + ibq + a\partial_q} ~ f(q) = e^{ic + iab/2} e^{ibq} e^{a\partial_q} ~f(q)= e^{ic + iab/2 +ibq}~ f(q+a), $$ for real coefficients a, b, c, the standard form used routinely in applications.
The first equality is a bland application of the Zassenhaus formula ("inverse Campbell-Baker-Haussdorf"), and the second equality is using up the Lagrange shift operator involved.