Let $K$ be an imaginary quadratic field. Let $\mathcal O$ be its ring of integers. Suppose that $2$ splits in $\mathcal O$, i.e. $2\mathcal O= \mathfrak p\overline {\mathfrak p} $ with $\mathfrak p$ and $\overline {\mathfrak p} $ not necessarily distinct prime ideals.
Let $k$ be a positive integer. What is the structure of the finite group $\mathcal O/2^k \mathcal O$?
The point about $2$ splitting as a product of two primes is not relevant here.
If a number field $K$ has degree $d$ and $\mathcal O_K$ is its ring of integers, then for any $n \in \mathbb N$, the additive group $\mathcal O_K / n \mathcal O_K$ is isomorphic to $\underbrace{\mathbb Z_n \oplus \dots \oplus \mathbb Z_n}_{d {\rm \ times}}$.
To see why this is, just consider an integral basis $\{ a_1, \dots, a_d \}$ for $\mathcal O_K$, and note that the elements $ m_1 a_1 + \dots + m_d a_d$ (with $m_1, \dots, m_d \in \mathbb Z_n$) form a set of representatives for the cosets of $\mathcal O_K$ modulo $n\mathcal O_K$.
In your example, $d = 2$ because $K$ is a quadratic field, so $\mathcal O_K / 2^k \mathcal O_K \cong \mathbb Z_{2^k} \oplus \mathbb Z_{2^k}$