Entanglement swapping is based on one trick: you can sum together states where Alice and Bob's qubits are entangled in different ways to create a system where their qubits are not entangled. It's possible to create such a state without having Alice and Bob interact with a third party (instead of each other), and the third party measuring their qubits will drop us into one of the entangled sub-cases.
Here's a mediocre diagram:
The left of the diagram shows a circuit for setting up the state shown on the right. The state diagram is arranged into a grid of local-to-Alice-and-Bob bits vs local-to-Eve bits. It shows the amplitudes for the sixteen possible qubit assignments (0000, 0001, 0010, 0011, 0100, etc). The amplitudes are shown as oriented circles, where right-facing is positive and left-facing is negative. For example: the top-left cell indicates the 0000 state having an amplitude of $+\frac{1}{4}$, the bottom-right cell indicates that the 1111 state has no amplitude, etc.
Anyways, imagine that you run some bell tests between Alice and Bob's qubits. They will fail the test, because in the overall system their qubits are not entangled. But if you throw out all the tests except the ones where Eve measured 00, putting us into the top row where $\psi_{AB} = \frac{1}{\sqrt{2}} \left|00\right\rangle + \frac{1}{\sqrt{2}} \left|11\right\rangle$, then that subset will be passing the tests!
The same is true of the other three Eve-measurement-outcome cases (01, 10, and 11): each violates the bell inequalities, but in different complementary ways that average into not passing any of the tests if you can't distinguish between the cases. (If you're allowed to perform operations conditioned on Eve's measurements before starting the bell tests, you can get all four cases into the same EPR state and pass unconditionally.)
Does that help?
There's no way to affect the observable behavior of the "upper" photons by manipulating the "lower" photons after the initial generation of the entangled pairs in the Glan-Thompson prism (if there were, you could use it to send a signal faster than light).
There's no interference pattern because the lower photons contain which-path information about the upper photons. Even if you never measure that information, its presence in the world prevents visible interference. There's no way to destroy the information except by reversing the process that originally created the entanglement, which is impossible in practice in this setup.
This answer has a more detailed discussion of a different version of this experiment. The key point is the same: if information about the light is preserved elsewhere in any form then there's no interference, and that information can't be and isn't destroyed, despite the name "quantum eraser".
Best Answer
The articles by Chris Monroe describe the following situations: (1) use of an already entangled pair of photons to transfer the entanglement, resulting in ion-ion entanglement; (2) interference of photons from independent sources, A, B, not previously entangled, such that no experiment can determine from which source a particular photon originated.
Method (2) typically relies on a beam splitter; Quantum interference of photon pairs from two remote trapped atomic ions shows this clearly, with the photons from the two independent sources being mixed together at the first beamsplitter, labeled BS.
The entanglement depends on lack of which-way information. The degree of entanglement acquired in this fashion may be less than that of down-converted photon pairs. Coincidence detection is required to keep members of a given pair temporally together.