There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since 
$$
\mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle,
$$
the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply
$$
T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger.
$$
This gives you the Kraus operators
$$
K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z.
$$