Let's start by finding a complementary channel for any channel given by a Kraus representation
$$
\Phi(X) = \sum_{k=1}^n A_k X A_k^{\dagger}.
$$
To make the necessary equations clear, let us assume that the channel has the form $\Phi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Y})$ for finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{Y}$. Let us also define $\mathcal{Z} = \mathbb{C}^n$; the complementary channel we will define will take the form $\Psi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Z})$. (For the channel in the question itself, we will have $\mathcal{X}$, $\mathcal{Y}$, and $\mathcal{Z}$ all equal to $\mathbb{C}^2$, but it helps nevertheless to assign different names to these spaces.)
Define an operator
$$
A = \sum_{k=1}^n A_k \otimes | k\rangle,
$$
which is a linear operator mapping $\mathcal{X}$ to $\mathcal{Y}\otimes\mathcal{Z}$. This gives us a Stinespring representation
$$
\Phi(X) = \operatorname{Tr}_{\mathcal{Z}} \bigl( A X A^{\dagger}\bigr).
$$
The channel
$$
\Psi(X) = \operatorname{Tr}_{\mathcal{Y}} \bigl( A X A^{\dagger}\bigr)
$$
is therefore complementary to $\Phi$. We can simplify this expression by observing that
$$
A X A^{\dagger} = \sum_{j=1}^n \sum_{k=1}^n A_j X A_k^{\dagger} \otimes | j \rangle \langle k |,
$$
so that
$$
\Psi(X) = \sum_{j=1}^n \sum_{k=1}^n \operatorname{Tr}\bigl(A_j X A_k^{\dagger}\bigr) | j \rangle \langle k |.
$$
There's not too much more we can do with this, except perhaps to use the cyclic property of the trace to obtain the expression
$$
\Psi(X) = \sum_{j=1}^n \sum_{k=1}^n \operatorname{Tr}\bigl(A_k^{\dagger} A_j X\bigr) | j \rangle \langle k |.
$$
Now let's plug in the specific operators from the question to obtain
$$
\Psi(X) = c_0^2 \operatorname{Tr}(X) | 0 \rangle \langle 0 | + c_1^2 \operatorname{Tr}(X) | 1 \rangle \langle 1 | + c_0 c_1 \operatorname{Tr}(\sigma_z X) | 0 \rangle \langle 1 | + c_0 c_1 \operatorname{Tr}(\sigma_z X) | 1 \rangle \langle 0 |.
$$
Here $\sigma_z$ denotes the Pauli-Z operator, which we get because $A_0^{\dagger} A_1 = A_1^{\dagger}A_0 = c_0 c_1 \sigma_z$. (I am assuming $c_0$ and $c_1$ are real numbers.) The expression may look a bit nicer in matrix form:
$$
\Psi\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\end{pmatrix} =
\begin{pmatrix} c_0^2(\alpha + \delta) & c_0 c_1 (\alpha - \delta)\\
c_0 c_1 (\alpha - \delta) & c_1^2 (\alpha + \delta) \end{pmatrix}.
$$
Finally, the question asks for Kraus operators of $\Psi$, which we can get by computing the Choi operator of $\Psi$. In general, this is the operator
$$
J(\Psi) = \sum_{j=1}^n\sum_{k=1}^n \Psi(|j\rangle\langle k|) \otimes |j\rangle\langle k|,
$$
and in this particular case we obtain
$$
J(\Psi) = \begin{pmatrix}
c_0^2 & 0 & c_0 c_1 & 0\\
0 & c_0^2 & 0 & -c_0 c_1 \\
c_0 c_1 & 0 & c_1^2 & 0\\
0 & -c_0 c_1 & 0 & c_1^2
\end{pmatrix}.
$$
This operator has rank 2, which means just 2 Kraus operators suffice. We can get them through a spectral decomposition of $J(\Psi)$. Specifically, we have
$$
J(\Psi) = \begin{pmatrix} c_0\\ 0\\ c_1\\ 0 \end{pmatrix}
\begin{pmatrix} c_0 & 0 & c_1 & 0 \end{pmatrix}
+ \begin{pmatrix} 0\\ c_0\\ 0\\ -c_1 \end{pmatrix}
\begin{pmatrix} 0 & c_0 & 0 & -c_1 \end{pmatrix},
$$
and by "folding up" these vectors we get Kraus operators:
$$
\Psi(X) = B_0 X B_0^{\dagger} + B_1 X B_1^{\dagger}
$$
where
$$
B_0 = \begin{pmatrix} c_0 & 0\\ c_1 & 0 \end{pmatrix}
\quad\text{and}\quad
B_1 = \begin{pmatrix} 0 & c_0 \\ 0 & -c_1 \end{pmatrix}.
$$