I actually have two different sub-questions, both based on the understanding of the quadratic term of the lagrangian, so the answer is probably linked. I will use the example of the linear-$\sigma$ model because it's the one I'm more comfortable with, so
Does the symmetry break, or is it either broken or unbroken?
With this I mean that I know that if $\mu^2>0$ the ground state is nondegenerate (so, no SSB) and if $\mu^2<0$ the ground state is degenerate (so, SSB). But does the value of $\mu^2$ change? Does my theory goes from being broken to be restored and vice-versa, or is every theory either broken OR unbroken, and then it stays that way? From what I read, it seems that SSB is a proerty of the theory, so it either is or isn't broken. But if this is the case, what does it mean for the electroweak theory to restore the symmetry at high temperatures? Is $\mu^2$ a function of $T$?
How does the mass come in?
From what I understood, the mass is defined as the quantity $m$ in the equation of motion: for the KG field $(\square-m^2)\phi=0$. Therefore, for the linear-$\sigma$ model, the mass should be the second derivative of $V(\phi_i^2)$ at the minimum, so if $\mu^2>0$ then $m=\mu$. If $\mu^2<0$, the value of the second derivative of $V$ at the minimum is different (because the minimum is different), so the mass in the broken case is different than the one in the restored case. Is my understanding correct? I'm doubtful because I also read that in the broken case, being $\mu^2<0$, the mass is not physical because $\mu$ isn't real.
These are the main questions that I have. There may be some deeper conceptual misunderstanding regarding the whole SSB idea: if you see some in the questions, please show it to me.