How is partial trace related to operator sum representation?

$|e_krangle$ is the basis of the environment. Taking the sum of projections onto an orthonormal basis of one subsystem is the definition of the partial trace over that subsystem.

I think it helps here to write things explicitly.

Suppose $mathcal E(rho)=operatorname{Tr}_E[U(rhootimes|mathbf e_0rangle!langlemathbf e_0|)U^dagger]$.

Pick a basis for the environment in which $|mathbf e_0rangle$ is the first element. Note that here $U$ is a unitary matrix in a bipartite system. The operator before taking the partial trace has matrix elements $$ [U(rhootimes|mathbf e_0rangle!langlemathbf e_0|)U^dagger]_{ij,kell} = sum_{alpha,gamma} U_{ij,alpha 0} rho_{alphagamma} (U^dagger)_{gamma0,kell} = sum_{alpha,gamma} U_{ij,alpha0}bar U_{kell,gamma0} rho_{alphagamma}. $$ Now notice that the partial trace amounts here to make $j=ell$ and sum over $j$ (because in this notation the indices $i,k$ refer to the system while $j,ell$ to the environment), so that we get $$ [mathcal E(rho)]_{ik} = sum_j [U(rhootimes|mathbf e_0rangle!langlemathbf e_0|)U^dagger]_{ij,kj} = sum_{alphagamma j} U_{ij,alpha0}bar U_{kj,gamma0}rho_{alphagamma}. $$ Notice how this is already essentially an operator sum representation: defining $(E_{j})_{ialpha}equiv U_{ij,alpha0}$, we get $$[mathcal E(rho)]_{ik}=sum_j (E_j)_{ialpha} (bar E_j)_{kgamma}rho_{alphagamma} = left[sum_j E_j rho E^dagger_jright]_{ik}.$$

Let ${|u_arangle}_{ain A}$ and ${|v_erangle}_{ein E}$ be orthonormal bases for the space $A$ and the environment space $E$ resp. Now, if we express $rho_A=sum_A alpha_a |u_arangle langle u_a|$ and $rho_E=|v_0rangle langle v_0|$ assuming that $rho_E$ is some pure state on the environment. Then we can write

$$ mathcal{E}(rho_A)=tr_Eleft(Uleft(rho_Aotimes rho_Eright)U^*right) = sum_{A}alpha_a, tr_Eleft(left(U|u_arangle |v_0rangleright)left(U|u_arangle |v_0rangleright)^*right) = sum_{A}alpha_a, sum_Elangle v_e|U|v_0rangle |u_arangle langle u_a|langle v_0|U^*|v_erangle =sum_Elangle v_e|U|v_0rangle left(sum_Aalpha_a|u_arangle langle u_a|right)langle v_0|U^*|v_erangle =sum_Elangle v_e|U|v_0rangle rho_Aleft(langle v_e|U|v_0rangleright)^*,$$

taking $E_e=langle v_e|U|v_0rangle$ be obtain the Kraus operators which act on the space $A$. I think its conceptually easier to see that the partial-trace unitary representation (or Stinespring representation) of a quantum channel is the purification of the Kraus representation, rather than the other way around.