'Unitary' charts on odd-dimensional spheres Consider the odd-dimensional sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$. One may talk variously about its structure as a contact, CR or Einstein-Sasaki manifold, but I'm looking for some specific down-to-earth detail that is hard to track down, namely charts that are complements of points and which are 'unitary' in the following sense.
The tangent space to $S^{n+1}$ at a point $x$ has a subspace isomorphic to the hermitian complement to the complex span of $x$. These patch together to give a complex vector bundle on the sphere -- and I'm fairly sure this is the holomorphic tangent bundle $HS^{2n+1}$ in the CR setting. Taking for $U$ the standard complement of the south pole, there is an isomorphism (even a CR isomorphism!) $\phi\colon Heis_n\simeq  U$, where $Heis_n$ is the Heisenberg group, diffeomorphic to $\mathbb{C}^n\times\mathbb{R}$. Hence we can consider the induced isomorphism $H\, Heis_n \to HU$ and ask whether it is unitary with respect to the standard hermitian structure on the left, and the one induced from $\mathbb{C}^{n+1}$ on the right. This is all controlled by the isomorphism $\phi$, and if I'm not mistaken in my calculations, taking standard stereographic projection rewritten in complex coordinates doesn't do the trick (which is, making the relevant adjustments, only orthogonal).

What's an explicit chart for an odd-dimensional sphere (considered as embedded in $\mathbb{C}^{n+1}$) satisfying the above condition?

Asking for such a chart (and analogously over the complement of the north pole) is equivalent to giving a local section of $SU(n+1) \to S^{2n+1}$ over $U$ (and $V$). Note that from a Riemannian point of view I want to consider the sphere with the homogeneous metric from the isomorphism $S^{2n+1} \simeq SU(n+1)/SU(n)$.
I have a hard time believing no one has written this down before.
 A: 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 
$$
|X_0|^2 = |X_1|^2 + \cdots +  |X_{n+1}|^2,
$$ 
by introducing homogeneous coordinates such that $z_i = X_i/X_0$.
Now one sees 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 $X_0-X_{n+1}=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,0,0,\ldots,0,1]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$.  To see this explicitly, make the coordinate change $X_0 = (Y+W)/2$, $X_{n+1}=(Y-W)/2$, leaving the others fixed, so that the equation for the hypersurface becomes
$$
\tfrac12(Y\overline{W}+W\overline{Y}) = |X_1|^2 + \cdots +  |X_n|^2,
$$
and now, on the complement of the hyperplane $W=0$, consider the affine coordinates
$$
u = \frac{Y}{W} = \frac{1+z_{n+1}}{1-z_{n+1}} \qquad\text{and}\qquad 
v_i = \frac{X_i}{W} = \frac{z_i}{1-z_{n+1}} \qquad 1\le i\le n, 
$$
Then the equation for the real hypersurface in these coordinates $(u,v_1,\ldots,v_n)$ on $\mathbb{C}^{n+1}$ is
$$
\mathrm{Re}(u) = |v_1|^2 + \cdots + |v_n|^2.
$$
This makes it easy to recognize as the Heisenberg model, and the formulae for $u,v_i$ in terms of the $z_i$ give the desired mapping.
In fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary' since all CR-isomorphisms between strictly pseudoconvex hypersurfaces are conformally unitary in the sense that the induced isomorphism between the corresponding holomorphic tangent spaces is always unitary up to a scalar multiple.
