Forgive me in advance, this may get overcomplicated. I am going to give you the facts as scaled down as I can but still sufficiently detailed. I think providing you with what we have and allowing you to infer from it is the best way to avoid misrepresenting the answer.
Here is the General Relativity equation that describes how gravity interacts with everything else:
$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi G_NT_{\mu\nu}$$
On the left side is gravity; it is described by terms that have to do with how space curves, expands, and contracts. On the right side is matter, radiation, etc. Pretty much all forms of energy.
It was determined through observation that the universe as we know it is expanding and that the rate of expansion is accelerating. To account for the acceleration of the expansion, the best-fitting theory includes a constant term:
$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}(R-2\Lambda)=8\pi G_NT_{\mu\nu}$$
Alternatively, one could choose to express that constant term on the right side of the equation. It makes no physical difference. Accordingly, you can interpret this term as a modification to how gravity affects spacetime (if added to the LHS) or as an additional energy term with a negative pressure (if added to the RHS).
Now, as per original GR, we have equations describing the expansion of the universe:
$$\frac{\ddot a}{a}=-\frac{4\pi G_N}{3}\sum_i(\rho_i+3p_i)$$
$a$ represents how much the universe has expanded, $\ddot a$ is the acceleration of the expansion, and $\rho_i$ and $p_i$ are the energy density and the pressure for the $i$th form of energy (the rest are constants). We put in matter, dark matter, radiation, and we even treat the background curvature of the universe as a form of energy here. What we find is that when we treat dark energy as an energy and set its pressure to be equal to negative of its energy density, our equations very closely match the observations. That is sufficient reason to like doing that. However, we also find (via a separate equation) that the energy density of something whose pressure equals the negative of the energy density remains constant for all time. This is unavoidable, it is one of the best fitting models so far.
Furthermore, the constant energy density presents other effects. The energy density of matter or radiation decreases with time. For instance, and this should be intuitive, the energy density of matter decreases like $a^{-3}$. That is, it drops like the cube of the expansion of the universe. And why not? as the universe expands, the volume increases like the expansion cubed; energy density is energy over volume, so it decreases like expansion cubed. Radiation goes like $a^{-4}$, and other energies decrease at varying rates. A constant density means that after a long time, it becomes the dominant term in the above equation. Effectively, after a long time:
$$\frac{\ddot a}{a}=\frac{8\pi G_N}{3}\rho_{DE}$$
This barrage of equations might mean nothing at all to you. That is fine. This is an answer to what dark energy is, why it has a constant energy density, and whether it has gravity.
As for whether or not dark energy can fall into a black hole, that is more complicated. From the point of view of modifying gravity, dark energy is not something that can fall into a black hole. However, from the point of view of being an additional energy term, one might think it must be able to fall into a black hole. Truthfully, I don't know that answer but I know it makes an insignificant difference. Dark energy is too weak to have an effect on the scale of black holes. Even the small gravity of the Sun is enough to negate the effect of dark energy throughout the solar system and probably a bit further. As you can see from the last equation, at late times the acceleration of the expansion is positive. Dark energy never brings the universe to a Big Crunch. For matter, radiation, etc, the acceleration terms at late times all are negative. Dark energy is preventing a Big Crunch.
In response to your last questions. This was never "proven". It was postulated decades ago and has been confirmed by many experiments but none that prove it beyond a doubt. As for what caused the Big Bang. The Big Bang was not an event, it was a moment of time. The Big Bang represents the point in time where the equation for $a$ (remember that's the term that represents how much the universe has expanded) goes to zero. That's it. It has no more need for a cause than has any moment of time after it.
Best Answer
Not really my field, but I'll take a crack at it.
Notice that the expression for the Jean's length $$ \lambda = \sqrt{\frac{k_B T r^3}{G M \mu}}$$ depends on both the size of the region in containing the mass and temperature of the cloud.
It should be obvious that the if a region of the cloud is contracting then $r$ gets smaller while $M$ remains the same. You need one more fact: gravitational potential energy is being converted into thermal energ (i.e. largely the kinetic energy of the particles, see also the Virial Theorem).
The wikipedia article takes it up from there "It is only when thermal energy is not equal to gravitational work that the cloud either expands and cools or contracts and warms, a process that continues until equilibrium is reached."
Notice *"until equilibrium is reached"**.
You've asked about the case from the wikipedia quote above where it contracts and warms. That said, if it gets warm enough it will lose energy relative it's surroundings by radiation, which means $T$ will drop and the cloud will again contract and warm. This continues until either the cloud stops behaving like a gas or the onset of fusion provides some new energy to keep the temperature up.
This is the case addressed above where it expands and cools.