Let $K$ be an imaginary quadratic field, $\mathcal{O}_K$ be its ring of integers, $\mathcal{O}$ be an order, $I_K, P_K$ be the group of ideals and principal ideals in $\mathcal{O}_K$, $I_K(m)$ be the $\mathcal{O}_K$-ideals primes to $m$, $I(\mathcal{O}), P(\mathcal{O})$ be the group of ideals and principal ideals in $\mathcal{O}$, $I(\mathcal{O},f),P(\mathcal{O},f)$ denote the $\mathcal{O}$-ideals and principal $\mathcal{O}$-ideals prime to $f$, where $f$ is the conductor of $\mathcal{O}$ and finally, let $C(\mathcal{O})$ denote the ideal class group of $\mathcal{O}$. Then, in the following Proposition,
Let $\mathcal{O}$ be an order of conductor $f$ in an imaginary quadratic field $K$. Then there are natural isomorphisms $$C(\mathcal{O}) \simeq I(\mathcal{O},f)/P(\mathcal{O},f) \simeq I_K(f)/P_{K,\mathbb{Z}}(f)$$ where $P_{K,\mathbb{Z}}(f)$ is the subgroup of $I_K(f)$ generated by principal ideals of the form $\alpha \mathcal{O}_K$, where $\alpha \in \mathcal{O}_K$ satisfies $\alpha \equiv a \mod f\mathcal{O}_K$ for an integer $a$ relatively prime to $f$.
is the integer $a$ a fixed integer that doesn't depend on $\alpha$ (which would mean that based on the $a$ we pick, we have the corresponding ideals in $P_{K,\mathbb{Z}}(f)$), or does it change with respect to each $\alpha$ (i.e., for every $a$, we pick corresponding $\alpha$'s and then run through all $a$ coprime to $f$)?
The $a$ depends on $\alpha$. The point of that last clause is only to exclude from $P_{K,\mathbb{Z}}(f)$ those principal ideals which are not relatively prime to $f$, because they have already been excluded from $I_K(f)$.