Now as for an intuitive explanation concerning Pete Clark's question:

I would be interested in, at least, a reference for the fact that
K(j(E),h(E[N])) contains the N-ray class field of K for arbitrary
orders.

Here is my intuitive point of view on it (I put it in another answer because I won't try to be perfectly rigorous), in terms of moduli.

Let $E_1$ be a an elliptic curve with maximal CM by $O_K$, and $E_2$ be a curve with CM by $O$, an order $O$ of conductor $F$ in $O_K$. $E_2$ is defined over $K(j(E_2)) \supset K(j(E_1))$, so there is a rational isogeny $E_2 \to E_1$ of degree $F$. If we add the field of definition of the points of $m$-torsion of $E_2$ then it is clear that if $F$ is prime to $m$, these points get transported to the points of $m$-torsion of $E_1$, so we already have the $m$-ray class field.

What is more surprising is that it works also when $m$ is not prime to $F$.
So let's assume that $m \mid F$ and see why we can still transport the $m$-torsion from $E_2$ to $E_1$.

The reason is as follow: when we are in the ring class field of $O$, all isogenies of degree $F$ between $E_1$ and an elliptic curve with endomorphism by $O$ are already rational; this means that the Galois action on $E_1[F]$ is given by a diagonal matrix. (And of course being in the $F$-ray class field means that the Galois action on $E_1[F]$ is the identity.) So in particular the kernel $K_1$ of the isogeny $E_1 \to E_2$ has a rational complement $K_2$.

Now if we have the $m$-torsion on $E_2$, pushing it through the dual isogeny $E_2 \to E_1$, we have that at least the points of $m$-torsion of $K_1$ are all rationals.
But because the points of $m$-torsion are rationals in $E_2$, the $m$-roots of unity are rationals, and so by looking at the Weil pairing we see that the points of $m$-torsion in $K_2$ are rationals. So all points of $m$-torsion in $E_1$ are rationals.

6more comments