What is the role of the vacuum expectation value in symmetry breaking and the generation of mass?

As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its nonvanishing vacuum expectation value cannot be given a particle interpretation. But the field $\psi=\phi-v$ has such an interpretation as its vacuum expectation value is zero. This works only if $v$ is the vacuum expectation of $\phi$.

Note that the field $\phi$ is and remains massless; it is the field $\psi$ that had acquired a mass term.

The 1-loop approximation to a quantum field theory is given by the saddle-point approximation of the functional integral. For that you have to expand around a stationary point, and for stability reasons this stationary point has to be a local minimizer. If the local minimum is not global, the vacuum state is metastable only; so one usually expands around the global minimizer.

A mass term breaks the scaling symmetry of a previously scale-invariant theory. It may or may not break other symmetries. In the above case, the symmetry $\phi\to-\phi~~~$ of the action is broken in the stable vacuum.

Firstly, note that 'particles' are quantised small oscillations of a field.

Now, to the potential. You would find a 'mass' term expanding around any point, but you'll also generically find a linear term. If you tried to perturbatively quantise the 'small oscillations' around such a point, you would find non-zero Feynman diagrams where particles are created from nothing. This represents an instability: the theory is telling you that you are doing the wrong thing.

Edit: I forgot to address the symmetry-breaking aspect. If you consider just the scalar field model, then the broken global symmetry means that you will have an exactly massless particle: the Goldstone boson. This corresponds to the angular oscillations (you only discussed the radial mode). In a gauge theory, a gauge boson is massless if and only if the vacuum preserves the corresponding symmetry.