**Part I:  The original question:**

Now that the question has been clarified, I can answer it.  The answer is 'no', there is no CR-isomorphism $\phi: \mathrm{Heis}\to U$ that is unitary on the holomorphic tangent bundles.

To see this, it's probably better to look at the dual $1$-forms.  Let $\alpha:T S^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $HS^{2n{+}1}\subset TS^{2n+1}$ and such that the $2$-form $\mathrm{d}\alpha$ restricts to $HS^{2n{+}1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $HS^{2n{+}1}$.  Similarly, let $\beta: T \mathrm{Heis}^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $H\mathrm{Heis}^{2n+1}\subset T\mathrm{Heis}^{2n+1}$ and such that the $2$-form $\mathrm{d}\beta$ restricts to $H\mathrm{Heis}^{2n+1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $H\mathrm{Heis}^{2n+1}$.  

If $\phi:\mathrm{Heis}\to U$ were a CR-isomorphism that was unitary on the corresponding holomorphic tangent bundles, one would necessarily have $\phi^*\alpha = \beta$.  However, because the integral of $\alpha\wedge(\mathrm{d}\alpha)^n$ over $S^{2n+1}$ is finite, its integral over $U$ is also finite.  On the other hand, the integral of $\beta\wedge(\mathrm{d}\beta)^n$ over $\mathrm{Heis}$ is infinite because it is the integral of a left-invariant volume form over a noncompact Lie group.  Thus, $\phi$ cannot exist.

By the way, it would not be hard to show, using the method of equivalence, that the inequivalence holds even locally, i.e., there is no open subset of $\mathrm{Heis}^{2n+1}$ that is unitarily CR-diffeomorphic to any open subset of $S^{2n+1}$.  However, that would take a little more argument, and I am not sure that the OP would be interested.

**Part II:  The modified question:**

Now that the OP has clarified (in the comments below) what is meant by 'comes from' and that it is not actually required that the CR-isomorphism be literally unitary, it's easy to see how to establish the desired CR-isomorphism:

First, it helps to realize the $(2n{+}1)$-sphere as a hypersurface in $\mathbb{CP}^{n+1}$ as the hypersurface
$$
|X_0|^2 = |X_1|^2 + \cdots +  |X_{n+1}|^2.
$$ 
Now it's clear that the group $\mathrm{SU}(1,n{+}1)\subset \mathrm{SL}(n{+}2,\mathbb{C})$ acts transitively on this hypersurface (and preserves the CR-structure, of course).  If you now 'de-projectivize' this by looking at the part of the hypersurface that lies in the in affine chart that is a complement to the hyperplane $X_0-X_1=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,1,0,\ldots,0]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$.  One can now show that there are complex coordinates on $\mathbb{C}^{n+1}$, say, $(u,v_1,\ldots,v_n)$ so that the image hypersurface is described by the equation
$$
\mathrm{Im}(u) = |v_1|^2 + \cdots + |v_n|^2.
$$

This makes it easy to recognize as the Heisenberg model and, in fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary'.