Calculating the effect of the second CNOT gate works pretty much as the first one; you just have to carry the state of the third qubit along (this you actually do in case of the first gate with the first qubit). Recall that everything here is linear – if you feel uncomfortable with such a manipulation of an entangled state, consider the effect of the gate on each term in the state and add them up.
Now, the calculation in some detail. After the first CNOT gate (i.e., at place 2) you have
$$ (\alpha|0\rangle+\beta|1\rangle)\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).$$
When the first qubit is in the state $|0\rangle$ nothing happens, when it is in the state $|1\rangle$, the second qubit is flipped, so that we have
$$ \frac{\beta}{\sqrt{2}}(|110\rangle+|101\rangle).$$
Combining the two terms gives the state at 3
$$\frac{1}{\sqrt{2}}[\alpha(|000\rangle+|011\rangle)+\beta(|110\rangle+|101\rangle)].$$
If you want to look at the state of just a single qubit which is entangled with other qubits, you have to trace out all the other qubits. You can look at the Wikipedia article on partial trace for more detail.
Finally, when you have a detector placed at the end you can see what happens by taking the scalar product of the state after the last Hadamard gate (i.e., the state at 4 that I will denote for now by $|\psi_4\rangle$) and the eigenstates of the measured observable. For example, when measuring $x$ [with the eigenstates $(|0\rangle\pm|1\rangle)/\sqrt{2}$] of the first qubit you calculate $\frac{1}{\sqrt{2}}(\langle 0|\pm\langle 1|)_1|\psi_4\rangle$ to see what state you get after detection of $(|0\rangle\pm|1\rangle)/\sqrt{2}$. (The subscript 1 means that the bra represents a state of the first qubit.) Similarly, you can proceed for any state and any measurement. Note that the state obtained in this way is not normalized; $|\frac{1}{\sqrt{2}}(\langle 0|\pm\langle 1|)_1|\psi_4\rangle|^2$ gives the probability of measuring $(|0\rangle\pm|1\rangle)/\sqrt{2}$.
Probably the most authoritative source on the Everett Interpretation that goes into the mathematics is Everett's thesis.
The way I understand it (and I can't claim to be an expert on this), the reasoning goes something like this.
Two non-interacting systems can be represented in a single joint equation like this.
$-i\hbar\frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & 0 \\ 0 & H_2}\pmatrix{\psi_1 \\ \psi_2}$
The matrix in the middle should be thought of as block-diagonal. Each block represents the (possibly infinite-dimensional) Hamiltonian of one of the systems. (Obviously, this is not a rigorous presentation...)
We now suppose that for a short period of time, the two sub-systems are allowed to interact. Interaction creates off-diagonal cross-terms in the Hamiltonian matrix in which the state of one system affects the other. (Because it's a Hamiltonian, when you integrate and exponentiate the results have to remain unitary - this is the 'unitary operator' I think you're looking for.)
$-i\hbar\frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & \epsilon_{12} \\ \epsilon_{21} & H_2}\pmatrix{\psi_1 \\ \psi_2}$
This interaction matrix can be diagonalised as $U^{-1}DU$ where $U$ is a constant unitary matrix of eigenstates, and $D$ is a diagonal (or at least block-diagonal) matrix. $D$ might have lots of eigenvalues (maybe infinitely many) but I'll show two to illustrate the idea.
$-i\hbar\frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=U^{-1}\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$
Move the $U$ across to the other side.
$-i\hbar\frac{\partial}{\partial t}U\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$
Now change variables to $\pmatrix{\phi_1 \\ \phi_2}=U\pmatrix{\psi_1 \\ \psi_2}$ to get:
$-i\hbar\frac{\partial}{\partial t}\pmatrix{\phi_1 \\ \phi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}\pmatrix{\phi_1 \\ \phi_2}$
This is again the equation for two non-interacting systems, but this time the two 'systems' in question are correlated linear combinations of the two original systems. Each represents a distinct outcome of the measurement; a separate 'world'.
(The division into eigenstates/worlds don't all have to be related to the states of the observed system. A lot of the choices will be to do with various quantum choices for events going on inside the observer, too. So it may be that for a two-choice quantum observable, what we get is two groups of $D_i$s, one group all corresponding to spin up, the other group to spin down. Again, I'm skipping past many details.)
If we are in the $\phi_1$ world, the state in the original basis will look like:
$U^{-1}\pmatrix{\phi_1 \\ 0}=\pmatrix{\psi_{11} \\ \psi_{21}}$
That's the observer in a state $\psi_{11}$ observing system 2 in state $\psi_{21}$. Similarly if we are in the $\phi_2$ world. The state of observer and observed are correlated in each 'world'.
That's my take on the idea. But for an authoritative answer, see Everett.
Best Answer
Shortest answer: No!
Short answer: No, in the many worlds interpretation measurement is described as an interaction between a "system to be measured" and a "measurement system". In quantum mechanics it is generally true that an interaction between 2 systems can lead to entanglement.
Longer answer: No, the many worlds interpretation does not distinguish between the entanglement of two systems under some interaction and "measurement". Measurement is simply the interaction between a system that is being measured and the measurement system. In general, but not always, this interaction can lead to entanglement between two system.
Whether the interaction leads to entanglement depends on the initial states of both the system and the measurement device AS WELL AS the particular form of the interaction Hamiltonian. I make this distinction because, when thinking about the many worlds interpretation, it is important to make sure you are making basis independent statements. For example, the initial state of your qubit might be $(1/\sqrt{2})(|0\rangle + |1\rangle)$ which is a superposition state in the z-basis, but is not a superposition state in the x basis. So, while tempting, we can NOT state the following conclusion: "Measurement of a superposition state leads to entanglement between the measurement system and measurement device". Something closer to the case might be: "When a system is in a superposition of the measurement basis (which is determined by the form of the interaction between the measured system and measurement device) the measurement leads to entanglement between the two systems.
edit to address:
Above I have claimed that in the Many Worlds Interpretation that measurement is no different from entanglement between the measurement device and measurement system.
But you seem to ask about the reverse. Is all entanglement a form of measurement?
The answer here is complicated, but, as I'll explain, doesn't really matter. What follows is my opinion. In my opinion, in the many worlds interpretation, a measurement arises when a measurement system interacts with a system to be measurement. The next question is then "what is a measurement system?" Intuitively there seems to be something at least qualitatively different between a qubit and a macroscopic photodetector. If we call the latter a measurement device and the former not a measurement device then, on my definition, entanglement with a photodetector DOES constitute a measurement while entanglement with a single qubit DOES NOT constitute a measurement.
This seems problematic because we are left with the problem that the Copenhagen interpretation faces which is "how do we define a measurement so that we know when the wavefunction collapses?" HOWEVER, there is a major difference. In the case of the Copenhagen interpretation the answer to the above question actually changes the rules of physics, whereas in the Many Worlds interpretation the answer to the question is purely semantic. The rules of physics do not change at all.
Above, I gave the opinion that entanglement with a qubit is not measurement while entanglement with a photodetector is. However, I'll state that, actually, in many cases, it CAN BE intuitively useful to think about entanglement with small quantum systems as a type of measurement. This is to say, in the Many Worlds interpretation, since it doesn't matter to the physics what is and what is not a measurement, we are free to move the definition of measurement (or the Heisenberg cut) around as much as we like. We can consider interaction with a qubit to BE a measurement or we can consider interaction with a human to NOT be a measurement. The physics is the same either way. This is the beauty and motivation of the many worlds interpretation.
Here are some of my (and your) other posts on the many worlds/everettian interpretation:
Finally a disclaimer: This post sounds like a glowing approval of Many Worlds interpretation. While it is my favorite interpretation, it is not without shortcomings. The main shortcoming is that it doesn't not prescribe any way to determine subject mental states from physical quantum states. This makes it impossible or very hard to answer the question: "What does a human experience when his brain is in a superposition state of having experience incompatible outcomes?" But at this point I'm off topic.