Throughout, I'll be using the Schrödinger picture and I won't take the most general approach (I'll also leave out some subtleties with duals and preduals, think of finite dimensional systems).
I would agree with the comments that you seem to be confusing a few notions here, but let me give it a shot: If you say "an operator, which takes us from system A to system B", you are talking about a map $T$ that (linearly, because we do quantum mechanics) sends density matrices to density matrices, i.e.:
Let $\mathcal{H}$ denote some Hilbert space, then a density matrix $\rho\in\mathcal{B}(\mathcal{H})$ (bounded linear operators on the Hilbert space - in finite dimensions those are just matrices) is a positive semidefinite matrix with unit trace ($\operatorname{tr}(\rho)=1$). Then "an operator which takes us from system A to system B" is a linear map
$$T:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}_B);~~~\rho_A\mapsto \rho_B:=T(\rho)$$
In order for that map to be a map that sends density matrices to density matrices, you must require that it is trace-preserving and completely positive (positive in the sense that it sends positive semidefinite matrices to positive semidefinite matrices, trace-preserving in that $\operatorname{tr}(T(\rho))=\operatorname{tr}(\rho)$ such that the density matrices are normalized and I'm going to skip the "complete" in "completely positive").
Incidentally, the time evolution defines such a map (which could have an additional parameter - the time). These maps that take you from one system to another don't have an expectation value (does the time evolution have an expectation value? No, you can only ask how the expectation of an operator changes over time - that however is a different question).
Now you talk about "expectation values". Those occur, if you measure a system (i.e. you measure an observable). A measurement doesn't take you from system A to system B, it takes the state and gives you a (real) value from some probability distribution. Granted, you can ask, how the state looks like after the measurement - and this then defines a map as above, but that is a different question. More precisely, given an observable $M$ and a state $\rho$, its expectation value is given by
$$ \langle M \rangle_{\rho}=\operatorname{tr}(\rho M) $$
as you write above. If you have a larger system that consists of multiple tensor factors, e.g. your system is $\mathcal{B}(\mathcal{H}_A)\otimes\mathcal{B}(\mathcal{H}_B)$ then your density matrices are positive, normalized elements $\rho\in\mathcal{B}(\mathcal{H}_A)\otimes\mathcal{B}(\mathcal{H}_B)$ and the above formula still holds. It might of course happen, that your state is a product state, i.e. $\rho=\rho_A\otimes \rho_B$, but then, the formula still holds and it just reads: $\operatorname{tr}(\rho M)=\operatorname{tr}(\rho_A\otimes \rho_B M)$. There is no difference if you have multiple tensor factors.
Let's now answer the question posed above: You have a state in system A, you send it to B, where you measure an observable $M\in\mathcal{B}(\mathcal{H}_B)$ - how would the expectation look like? Well, if $T:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}_B)$ is the map that sends us from A to B and our state is $\rho\in\mathcal{B}(\mathcal{H}_A)$, then
$$\langle M\rangle_{T(\rho)}=\operatorname{tr}(T(\rho)M)$$
Maybe that already clears up enough for you?