= Solution
Here a divisor is a <modulus of a number field>, rather than an arbitrary real-weighted divisor. Write
$$
\mathfrak c=\mathfrak c_0\mathfrak c_\infty,\qquad \mathfrak c_0=\prod_{\mathfrak p}\mathfrak p^{m_{\mathfrak p}},
$$
where $\mathfrak c_0$ is a nonzero <integral ideal>, the finite multiplicities $m_{\mathfrak p}$ are nonnegative integers, and $\mathfrak c_\infty$ is a set of real <Archimedean places>. The <multiplicity of a place in a modulus> is $m_v(\mathfrak c)=m_{\mathfrak p}$ at the finite place corresponding to $\mathfrak p$, is $1$ at a real place in $\mathfrak c_\infty$, and is $0$ at every other infinite place. In particular complex places have multiplicity $0$.
Let $I_{\mathfrak c}$ be the group of nonzero <fractional ideals> prime to $\mathfrak c_0$. Put
$$
K_{\mathfrak c,1}^{\times}=\{a\in K^{\times}:v_{\mathfrak p}(a-1)\geq m_{\mathfrak p}\text{ for }\mathfrak p\mid\mathfrak c_0,\quad \sigma_v(a)>0\text{ for }v\in\mathfrak c_\infty\}.
$$
Let $P_{\mathfrak c}$ consist of the <principal ideals> $(a)$ with $a\in K_{\mathfrak c,1}^{\times}$. The <generalized ideal class group> is the <ray class group>
$$
\boxed{H_{\mathfrak c}=I_{\mathfrak c}/P_{\mathfrak c}.}
$$
Let $U=\mathcal O_K^{\times}$ be the <unit group>, let $U_{\mathfrak c}=U\cap K_{\mathfrak c,1}^{\times}$ be the group of <ray units>, and define the <residue and signature group of a modulus>
$$
G_{\mathfrak c}=(\mathcal O_K/\mathfrak c_0)^{\times}\times\{\pm1\}^{\mathfrak c_\infty}.
$$
The first factor is omitted when $\mathfrak c_0=\mathcal O_K$. There is a <group homomorphism> $\rho:U\to G_{\mathfrak c}$ recording the unit's finite residues and its signs. Its <kernel> is $U_{\mathfrak c}$.
Let $P^{\mathfrak c}$ consist of all <principal fractional ideals> prime to $\mathfrak c_0$, and put $R_{\mathfrak c}=P^{\mathfrak c}/P_{\mathfrak c}$. The two <short exact sequences> are
$$
\boxed{1\longrightarrow U/U_{\mathfrak c}\xrightarrow{\rho}G_{\mathfrak c}\longrightarrow R_{\mathfrak c}\longrightarrow1,}
$$
and
$$
\boxed{1\longrightarrow R_{\mathfrak c}\longrightarrow H_{\mathfrak c}\longrightarrow\operatorname{Cl}(\mathcal O_K)\longrightarrow1.}
$$
For the first <short exact sequence>, <weak approximation for number fields> realizes every choice of finite unit residues and real signs by some $a\in K^{\times}$ prime to $\mathfrak c_0$. Send that data to the class of $(a)$ in $R_{\mathfrak c}$. Changing $a$ without changing its residues and signs multiplies it by an element of $K_{\mathfrak c,1}^{\times}$, so the map is well-defined. Its <kernel> consists exactly of data arising from units: if $(a)=(b)$ with $b\in K_{\mathfrak c,1}^{\times}$, then $a/b\in U$. This gives $R_{\mathfrak c}\cong G_{\mathfrak c}/\rho(U)$. For the second <short exact sequence>, forget the ray conditions. Its <kernel> is $R_{\mathfrak c}$, and <weak approximation for number fields> gives a representative prime to $\mathfrak c_0$ for every <ideal class>. These are the two parts of the <ray class exact sequence>.
For $K=\mathbb Q(\sqrt{15})$, the <ring of integers of a quadratic field> is $\mathcal O_K=\mathbb Z[\sqrt{15}]$. The finite modulus is trivial and both real places occur, so $H_{\mathfrak c}$ is the <narrow ideal class group> and $G_{\mathfrak c}=\{\pm1\}^2$. The two <field embeddings> send $\sqrt{15}$ to $\sqrt{15}$ and $-\sqrt{15}$. The given unit $\epsilon=4+\sqrt{15}$ is positive at both places, since $4-\sqrt{15}>0$; the unit $-1$ is negative at both. Thus the <unit signature map> has image
$$
\rho(U)=\{(+,+),(-,-)\},\qquad |R_{\mathfrak c}|=\frac42=2.
$$
The given <ideal class group> has order $2$. The second <short exact sequence> therefore gives
$$
\boxed{|H_{\mathfrak c}|=2\cdot2=4.}
$$
To determine the group structure, retain the ideal $\mathfrak p=(2,1+\sqrt{15})$ that generates the ordinary <ideal class group>. We have $\mathfrak p^2=(2)$: all generators $4$, $2(1+\sqrt{15})$ and $(1+\sqrt{15})^2$ lie in $(2)$, while
$$
(1+\sqrt{15})^2-2(1+\sqrt{15})-3\cdot4=2
$$
puts $2$ in $\mathfrak p^2$. Since $2$ is <totally positive>, $[\mathfrak p]^2=1$ in the <narrow ideal class group>. Its image in the ordinary <ideal class group> is nontrivial, so $[\mathfrak p]$ has order exactly $2$.
The class of the <principal ideal> $(\sqrt{15})$ is a nontrivial element of $R_{\mathfrak c}$. Its two signs are $(+,-)$, and multiplying by a unit can only reverse both signs or neither, so no generator of this ideal is <totally positive>. Its square is $(15)$, which does have a <totally positive> generator. Thus $[(\sqrt{15})]$ is another element of order $2$, distinct from $[\mathfrak p]$ because its ordinary <ideal class> is trivial. These two elements are independent and generate all four classes. Consequently
$$
\boxed{H_{\mathfrak c}\cong(\mathbb Z/2\mathbb Z)^2.}
$$
The nontrivial ordinary <ideal class> already has a lift of order $2$, so the extension in the second <short exact sequence> splits; it cannot be cyclic of order $4$.
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