1&2 are in a Bell state as well as 3&4, i.e., each of these pairs is maximally entangled. There are no correlations between 2&3 or 1&4 in the initial state. I haven't claimed anything like that.
Now if you project to a Bell state of 2&3, in the so prepared subensemble 2&3 are fully entangled, because they are prepared to be in a Bell state. Due to the entanglement of 1&2 and 3&4 this implies that for this subensemble also 1&4 are in a Bell state.
It's a easy, although some elaborate calculation, as described by Jennewein et al in
https://arxiv.org/abs/quant-ph/0201134
It's more easily seen in the 2nd-quantization notation. The four Bell states of a photon pair with momentum labels ##j## and ##k## are created from the Vakuum by
$$\hat{\Psi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,V) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,H)],$$
$$\hat{\Phi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,H) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,V)].$$
Then the initial four-photon state can then be written in two forms,
$$|\Psi_{1234} \rangle = \hat{\Psi}_{12}^{\dagger -} \hat{\Psi}_{34}^{\dagger-} |\Omega \rangle,$$
but this can as well be written as
$$|\Psi_{1234} \rangle=\frac{1}{2} (\hat{\Psi}_{23}^{\dagger +} \hat{\Psi}_{14}^{\dagger +} - \hat{\Psi}_{23}^{\dagger -} \hat{\Psi}_{14}^{\dagger -}-\hat{\Phi}_{23}^{\dagger +} \hat{\Phi}_{14}^{\dagger +} + \hat{\Phi}_{23}^{\dagger -} \hat{\Phi}_{14}^{\dagger -})|\Omega \rangle.$$
The former notation shows that photon pairs (12) and (34) are each in the polarization-singlet Bell state but (14) and (23), i.e, are uncorrelated.
The latter notation shows that if you project pair (23) to either of the four Bell state the pair (14) must be found in the same Bell state. In Pan et al's work, which we discuss here, (23) has been projected to the polarization-singlet state, and it has been demonstrated that then also the pair (14) is then in the same polarization-singlet state. This happens with probability 1/4.