I will assume familiarity with bra-ket notation, which makes this much easier to explain.
Schematically, what a single type-1 SPDC crystal does is it takes photons that are polarized along one axis and splits them into two photons of lower energy and the opposite polarization. Photons that are polarized along the perpendicular axis just pass through the crystal undisturbed. So, if the downconversion axis of the crystal is labelled 'V' and the other axis is 'H' (for vertical and horizontally polarized photons), this corresponds to the following rules:
|V_p\rangle\rightarrow |H_1 H_2 \rangle$$
Here 'p' labels the 'pump' photons, which are just the laser photons, and 1,2 label the two photons of lower energy that a laser photon splits into.
(By the way, just for completion I should mention two complications: this splitting process actually only happens very rarely for a polarization aligned with the downconversion axis, and in general you can also have crystals where one V changes into two Hs. Neither of these are relevant for our purposes.)
This means, using the linearity of quantum mechanics, that a superposition state with both horizonal and vertical components will evolve into the following state:
$$\alpha|H_p\rangle+\beta|V_p\rangle \rightarrow \alpha|H_p\rangle+\beta|H_1 H_2\rangle$$
So the photon ends up in a superposition of being downconverted and just passing through. This is somewhat interesting, but it does not have any polarization entanglement. In particular, if you select for only photons that are downconverted, which is what happens in experiments in practice (you use filters to get rid of all the photons at the pump frequency), you end up with the clearly separable state $|H_1 H_2\rangle$.
However, if you add another SPDC crystal that is right after the first one and oriented at 90 degrees relative to it, it will do downconversion according to the rules
|V_p\rangle\rightarrow |V_p \rangle$$
$$|H_p\rangle\rightarrow |V_1 V_2\rangle
since the downconversion axis is now along H. And this means that the initial pump photon in a general state will go through the following transformations:
$$\alpha|H_p\rangle+\beta|V_p\rangle \rightarrow \alpha|H_p\rangle+\beta|H_1 H_2\rangle \rightarrow \alpha|V_1 V_2\rangle+\beta|H_1 H_2\rangle$$
where we are ignoring the very tiny probability of any nonlinear process happening to the $|H_1 H_2\rangle$ photons as they pass through the second crystal. This is an entangled state. It is maximally entangled when $|\alpha|=|\beta|$, which conveniently just corresponds to an initial polarization that is equally between H and V, which is to say diagonal. So, without considering the various experimental complications that can limit this in practice, all you need to do is shine diagonal linearly polarized light through the two BBO crystals and out come the maximally entangled downconverted photons.