How to measure a multipartite density matrix?

I consider the tensor product $H_1 otimes H_2$ for pairs of qubits.
they are described by a 4*4 density matrix $rho$.
Bob and Alice share them and measure them randomly in many directions.
At the end they meet and try to calculate the elements of the density matrix (using a vertical direction for the basis).
One way to calculate $rho$ is to use 16 (or $N^2$ in the general case) hermitian orthonormal operators $Omega_i$:
$$rho = s_1 Omega_{1} + …. + s_{16} Omega_{16} $$ in an unique decomposition.
This is the case if they are orthonormal in the trace sense:
$$Tr( Omega_i Omega_j) = delta_{i,j}$$
According with this, if we take the mean value of $Omega_i$ in the density matrix state we compute
$Tr(rho Omega_i)$ and the result is $s_i$ . So the density matrix is given
by 16 mean values.
One can find these 16 operators like that

1. Take a null 4*4 density matrix and get 4 operators by placing one 1 on the diagonal.

2. Take your null matrix and place one 1 above the diagonal (6 possibilities)
   and place another 1 symmetric to the former with the diagonal.

3. Do the same thing with a i above the diagonal and a symmetric -i under the diagonal (6 other operators)

We have the 16 hermitian orthonormal operators. $N^2$ in the general case

I see that for the operator with a 1 in the off diagonal corners dd + uu is an eigenvector (eigenvalue = 1) and dd – uu also with -1 as eigenvalue.
The operator with an i in the upper right corner has dd – iuu as eigenvector with eigenvalue 1, and dd + iuu with eigenvalue -1.
the mean values will be between -1 and +1.
What are the results of Bob and Alice that i need for each of the 16 mean values? And how do i calculate this mean value.

I see that the 12 $Omega_{jk}$ (j not equal to k) are symmetries with respect to a direction in $H_1 otimes H_2$
the 4 others being projectors on a direction.

Consider the density matrix
(|00><00| + |11><11|)/2 + (|00><11| + |11><00|)/3
How will Bob and Alice get this term with 1/3?
Of course a general answer would be great.