All these statements are meant to be understood in terms of large populations of particles. For example, if the mean energy per particle in the Universe is smaller than the mass of electron, it is impossible to produce many of them and they in general will annihilate or transform into other species. In thermal equilibrium this can be understood just by looking at the statistical weight of electrons — it is proportional to $e^{-m/T}$ which makes their population negligible. Electrons still exist, it is just there are never cosmologically many of them.
The other thing is that in cosmology not everything can achieve equilibrium. Equilibration requires time and the expansion of the Universe actively counteracts it by, basically, reducing the rates of interactions. Freeze out represents the case when interaction rate of the particle is smaller than the Hubble rate. In this case, reactions effectively cease and population of these particles as a whole changes only due to expansion of the space.
I recommend to read Kolb, Turner "The Early Universe" to find out more.
I think I should use an example in my lecture note to avoid those questions that are too complicated. Let me see if I get this right. In the neutrino decoupling case, it should be $ν+\overline{ν}↔e^++e^−$ when it is thermally equilibrium. After the temperature drops below the energy of electron, there is no sufficient energy for $ν+\overline{ν}→e^++e^−$ to happen. So only $e^++e^-→ν+\overline{ν}$ in the universe. So neutrinos propagates freely after since. Is the frozen out particle neutrino?
No, you are wrong. What makes you think that the reactions you show are the only ones happening in the plasma? What about $\nu+\nu \leftrightarrow \nu+\nu$ and many others?
You should to think about this in the following way.
In the past, the Universe was dense and hot and particles could reach both dynamical and chemical equilibrium (look them up, the difference is important!)
We are for some reason interested in temperatures around $1 MeV \approx 2 m_e$. Are electrons in equilibrium then? In the expanding Universe they are if
$$ H \ll \Gamma_e = \langle n \sigma v \rangle $$
where $\Gamma_e$ is rate of some reaction involving electrons. If all reactions are much faster than Hubble rate, electron is definitely in equilibrium. If none — than electron decouples. In the case when some are and some are not — it is hard to tell in general, but sometimes we can.
What reactions do we have for electrons? Hundreds. There are at least to prominent groups — weak reactions (e.g. $\nu+e \to \nu +e$) and electromagnetic reactions ($e + \gamma \to e + \gamma$). Luckily, the rate of electromagnetic reactions is really huge so electrons are definitely in equilibrium.
Cool, then what will happen with them? Well, they will follow the Boltzmann distribution $\propto e^{-\frac{m}{T}}$ and quickly disappear with the falling temperature.
And that's it? Actually, no. There is a small asymmetry in electrons related to baryon asymmetry and electro-neutrality of the Universe, so about 1 in a billion electrons will survive, because it won't find a positron to annihilate with.
Right. And what about neutrinos? They are not charged, they don't interact electromagnetically. Actually, a dimensional estimate of the weak rate ($\Gamma \propto G_f^2 T^5$) gives us that weak reactions rate becomes smaller than the Hubble rate at temperatures about $3 MeV$. So neutrinos will freeze-out.
And neutrons? Not charged and interact weakly as well. The same thing, neutrons freeze-out a bit later at $\approx 1 MeV$ (which is related to their compound nature) and propagate freely. But they still decay, so if nothing happens, they would completely turn into protons in an hour or so.
Does something happen to prevent this? Why, yes — it is Big Bang Nucleosynthesis. But this is a story for other time. (see here and here)
Adding any more details is waste of time as I am already giving you a synopsis of every book on cosmology and particle physics.
