For any orthonormal basis that you pick, call it $|e_i\rangle$, you can write a matrix in terms of that basis as $$ \rho=\sum_{i,j}\rho_{i,j}|e_i\rangle\langle e_j|. $$ When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write $$ |e_i\rangle=|n\rangle|\nu\rangle, $$ splitting the sum over $i$ into two sums, over $n$ and $\nu$. Hence, you can write $$ \rho=\sum_{n,\nu,m,\mu}\rho_{n,\nu,m,\mu}|n\rangle\langle m|\otimes|\nu\rangle\langle\mu|. $$ It just happens that they've chosen to change where the different indices are put.
As for the partial transpose, that is simply the definition of the partial transpose. It may not seem so familiar, but recall with the transpose does, $$ \langle e_i|\rho|e_j\rangle=\langle e_j|\rho^T|e_i\rangle. $$ All the partial transpose does is that it only switches the left and right indices for the second subsystem, not the first.
In fact, this definition of $$ \langle n,\nu|\rho|n,\mu\rangle=\langle n,\mu|\rho^{T_B}|m,\nu\rangle $$$$ \langle n,\nu|\rho|m,\mu\rangle=\langle n,\mu|\rho^{T_B}|m,\nu\rangle $$ is a general definition. It's not about whether $\rho$ is separable or not. The point is that if it is separable, then $\rho^{T_B}$ is still a valid quantum state. On the other hand, if $\rho$ were not separable, there is no reason why $\rho^{T_B}$ need be a valid quantum state. In particular, it could have negative eigenvalues.
