It is possible to see the cotangent bundle of the submanifold as a kind of symplectic reduction of the cotangent bundle of the ambient manifold. I think it might be enough to explain the analogous fact from linear algebra.
Let V be a vector space and U a subspace. There is a natural symplectic form $\omega_V$ on $V^*\oplus V$ given by
$$
\omega_V((\alpha,u),(\beta,v)) = \alpha(v) - \beta(u)
$$
where greek letters are elements of $V^*$ and roman letters are elements of $V$. (This is just d of the Louville form in this situation.) There is an analogous form $\omega_U$ on $U^* \oplus U$.
Now, let $U^0$ denote the annahilator of $U$ in $V^*$. Consider the subspace
$$
U^0 \times\{0\} \subset V^* \oplus V
$$
This subspace is isotropic for $\omega_V$. Its symplectic complement is the coisotropic subspace $V^*\oplus U$.
Now it is a standard fact in symplectic geometry that if you divide a coisotropic subspace by its symplectic complement the result is naturally a symplectic vector space. (This is the linear algebra behind symplectic reduction.) Applying this idea here we see that the quotient
$$
(V^*\oplus U )/ (U^0\times\{0\})
$$
inherits a natural symplectic structure. Of course, the quotient is precisely $U^*\oplus U$ and the symplectic form is nothing but $\omega_U$.
Disclaimer: I don't talk to people about orbifolds. This answer may not represent the opinions of orbifolders.
As I understand it, the orbifold $\mathbb R^n/G$ is characterized by how manifolds map to it,† not by how it maps to manifolds. In particular, the orbifold is not determined by the ring of smooth functions on it (i.e. the ring of $G$-invariant smooth functions on $\mathbb R^n$). In particular, the ring of 3-fold symmetric smooth functions on $\mathbb R^2$ is isomorphic to the ring of 4-fold symmetric functions on $\mathbb R^2$.
Given that the smooth functions on an orbifold don't "remember" everything about it, it seems unreasonable to define the tangent space at a point in terms of derivations of smooth functions. However, we have another construction of the tangent space at a point: equivalence classes of smooth curves through that point. Since this definition has to do with maps into the space rather than maps out of it, we expect it to play well with orbifolds. Any smooth curve through the cone point of $\mathbb R^n/G$ lifts to a curve in $\mathbb R^n$, and any curve in $\mathbb R^n$ induces a curve in $\mathbb R^n/G$, so the tangent space to $\mathbb R^n/G$ should be the same as the tangent space to $\mathbb R^n$. Well, not exactly, since a curve in $\mathbb R^n/G$ can lift to a curve in $\mathbb R^n$ in $G$ different ways. So the tangent space to $\mathbb R^n/G$ at the cone point should really be the quotient of the tangent space of $\mathbb R^n$ by the action of $G$.
So far, it seems like I'm arguing that the tangent space to $\mathbb R^n/G$ at the cone point should be the orbifold $\mathbb R^n/G$. But I actually want to say that the tangent space should be the tangent space of $\mathbb R^n$, together with the action of $G$. The reasoning is that the vector space together with the action of the residual group is independent of how you express $\mathbb R^n/G$ as a quotient by a finite group. In other words, even though an isomorphism of orbifolds $\mathbb R^n/G\cong M/G'$ does not induce isomorphisms $\mathbb R^n\cong M$ or $G\cong G'$, it does induce isomorphisms of tangent spaces and residual groups in such a way that respects the actions of the residual groups on the tangent spaces. Once your definition of tangent space is canonically related to the orbifold, you should be welcome to think of it however you like.
† Roughly, a map from a manifold $M$ to $\mathbb R^n/G$ should be the same thing as a map from $M$ to $\mathbb R^n$, except that maps that differ by the action of $G$ should be regarded as the same map. More precisely, I think a map from $M$ to $\mathbb R^n/G$ should consist of a $G$-fold covering space of $M$ with a $G$-equivariant map to $\mathbb R^n$.
Best Answer
CR does stand for Cauchy-Riemann.
CR structures on 3 dimensional manifolds arise as the boundaries of complex (or almost-complex) 4 manifolds; if these boundaries are strictly pseudo-convex (i.e. convex in "holomorphic directions") the CR structure on the 3-manifold is a contact structure (if the boundary is only (pseudo-)convex or (Levi) flat, the CR structure integrates to a confoliation or a foliation respectively). There can be infinite dimensional families of foliations on a 3-manifold; more generally, whenever the CR structure is "non-generic" or integrable, one has continuous moduli, otherwise (eg in the contact structure case) one has discrete moduli (to be explicit: what has discrete moduli is the contact structure, not the "CR+contact structure".)