I'll try: Let's say you've got two systems, [itex]A[/itex] and [itex]B[/itex], each of which can be in the states [itex]|0\rangle[/itex] or [itex]|1\rangle[/itex] (perhaps think of a fermion with spin up/down states), or of course in any superposition of both, [itex]\alpha|0\rangle + \beta|1\rangle[/itex], with [itex]|\alpha|^2 + |\beta|^2=1[/itex]. The combined system then can be in any of the states [itex]|\psi\rangle_{AB}=\sum_{i,j}c_{ij}|i\rangle_A \otimes |j\rangle_B[/itex]. If this state can be written in the form [itex]|\psi\rangle_A \otimes |\psi\rangle_B[/itex], it is called separable; if not, it is entangled.
An example of an entangled state is [itex]|\Psi^-\rangle = \frac{1}{\sqrt{2}}(|0\rangle_A \otimes |1\rangle_B - |1\rangle_A \otimes |0\rangle_B)[/itex]. Its entangled nature comes to light if we let [itex]A[/itex], conventionally called Alice, perform a measurement. If she obtains the outcome 0 (with probability [itex](\frac{1}{\sqrt{2}})^2 = \frac{1}{2}[/itex], the state collapses to [itex]|0\rangle_A \otimes |1\rangle_B[/itex], and we know with certainty that [itex]B[/itex] (Bob) will obtain 1 upon measuring the system; conversely, if Alice obtains 1, the state afterwards will be [itex]|1\rangle_A \otimes |0\rangle_B[/itex], and thus, Bob's subsequent measurement will yield 0 as a result.
Now, the state [itex]|\Psi^-\rangle[/itex] is what's called pure, which basically means that it can be represented by a unique ray in Hilbert space (i.e. a single ket vector [itex]|\psi\rangle[/itex]). The converse of pure is mixed. A state is mixed if it consists of an ensemble of pure states -- you can picture this as being uncertain about what state the system is actually in. So if you have an apparatus with a randomizing element that prepares you state [itex]|\psi_1\rangle[/itex] with probability [itex]p_1[/itex], and state [itex]|\psi_2\rangle[/itex] with probability [itex]p_2[/itex] (such that [itex]p_1 + p_2 =1[/itex]), you describe whatever comes out of the apparatus by the statistical mixture of these two states.
Unfortunately, the bra-ket formalism is not well suited to the description of mixed states; to do so, one typically turns to the density matrix formalism. For a mixture of states such as the one above, the density matrix is: [itex]\rho = \sum_i p_i|\psi_i\rangle \langle \psi_i|[/itex]; it gives the probability with which the system is found in either of the states [itex]|\psi_i\rangle[/itex].
Now, a consequence of entanglement is that you can't associate to either of the systems [itex]A[/itex] or [itex]B[/itex] a pure state anymore. This is intuitive -- because of the entanglement, the systems considered on their own do not describe the complete state. Rather, the state of the system [itex]A[/itex] is described by the partial trace over [itex]B[/itex] of the density matrix of the whole system: [itex]\rho_A = tr_B \rho_{AB}[/itex] (nevermind the mathematical terminology; this just means 'whatever's left over when I forget about all the degrees of freedom associated to [itex]B[/itex]).
The last little fact we need is that pure states have zero entropy, while the entropy of mixed states is always nonvanishing. So effectively, if I have an entangled state as above, and restrict my attention to one part of it (say one particle of an entangled two-particle system), then I must describe the state of that part as having nonzero entropy, even though the complete entangled state has no entropy. Because the situation is completely symmetric, the entropy of one part is equal to the entropy of the other, if I remove (say, hide behind a horizon) the remaining one. I.e. if [itex]S[/itex] denotes entropy, [itex]S(A)=S(B)[/itex].
This almost directly leads to the 'area law' scaling of entanglement entropy: if I have some volume uniformly filled with some field, and remove a (spherical, for convenience) portion of it, then the entropy of the removed part relatively to the rest must be equal to the entropy of the rest relatively to the removed part (from inside the sphere, effectively the rest of the universe has been hidden behind the 'horizon'); but the area of the sphere's boundary is the only quantity both sectors have in common, so the entropy must end up proportional to it. (Vacuum correlations are just the correlations -- i.e. entanglement -- that are naturally present in the field.) Unfortunately, while the Bekenstein-Hawking entropy has a definite upper bound, given by the Planck area, the entanglement entropy hasn't -- I can always go to smaller and smaller distances and find higher and higher modes that contribute. What Jacobson's now claiming, essentially, is that gravity, which emerges from the thermodynamics of the horizon (recall, what has entropy, also has temperature), serves to regulate this divergence (if I understand correctly).
Does this help?