Short answer: it is a combination of (1) the ignition occurring in an electron-degenerate, isothermal core in which the equation of state is independent of temperature; and (2) the extreme temperature dependence of the triple alpha He fusion reaction.
Details:
The helium flash occurs at the tip of the first ascent red giant branch in stars with masses between 0.5 and about 2 solar masses. At this point the star consists of a helium core surrounded by a vigorously burning shell of hydrogen, surrounded by a very large convective envelope.
The core is the left over from core hydrogen burning, supplemented by helium produced by the hydrogen shell burning that takes over once the core hydrogen is exhausted. The inert core shrinks in radius from its main sequence size because it has more mass per particle, so must increase in density to maintain pressure. As it does so, the virial theorem demands that it also gets hotter. The shell burning drops more and more He into the core, the core shrinks further and gets hotter.
In stars > 2 solar masses, the core gets hot enough to ignite helium in the triple alpha process. This raises the core temperature, but not massively, because at the same time, the pressure increases, the core expands vigorously and the hydrogen burning shell is pushed outwards and extinguished.
In a lower mass star it is different. The He core density rises to the point that the core electrons become degenerate. Electron degeneracy pressure (EDP) dominates the total pressure of the gas and arises because at high enough densities, the electrons fill all low energy quantum states.EDP only depends on density, not temperature.
A core supported by EDP gets smaller and denser, the more massive it is, so becomes more degenerate as He ash is dropped onto it. But it is also sitting inside a hugely luminous H burning shell which heats it. In stars more massive than 0.5 solar masses, eventually the core gets hot enough ($\sim 10^{8}$ K) to ignite the He. Because degenerate electrons are extremely conductive, the core is almost isothermal, so ignition spreads rapidly through the core. This raises the core temperature, but crucially, not the core pressure (EDP is independent of temperature and, as Ken G explains in his answer, most of the heat gets absorbed by the non-degenerate ions that hardly contribute to the pressure). It just gets hotter and He fusion increases massively because it is very temperature sensitive (roughly proportional to $T^{40}$ !!). This runaway process is termed the "helium flash".
Eventually the temperature rises enough (to about $3\times10^{8}$ K) to break the electron degeneracy, the core expands rapidly, the H shell is extinguished and the core luminosity falls.
No, the Sun is not thought to have formed around a solid core, and solids would not exist at the temperatures and pressures at the centre of the protosun. The Sun formed simply from the gravitational collapse of a large cloud of gas.
The situation for Jupiter is different because far out in the circumstellar disc of the forming solar system, it was cool enough for the condensation of solids. The core accretion model is where gas giant planets begin their lives through the growth of a small (well maybe 10 Earth masses!) rocky/icy core, this is later followed by a brief and rapid accretion of the gaseous envelope.
It is currently not known whether Jupiter has a rocky core or not. That is one of the key questions that it is hoped the Juno mission will answer.
Recent years have seen the re-emergence of the thermal instability model for the rapid formation of gas giants. Such giants would not have a solid core and this formation mechanism is more akin to the way that the Sun formed.
As to whether there could be observational evidence that the Sun did not form around a solid core, I am doubtful. A solid core would of course vapourise at the high temperatures inside the Sun, but its enhanced mean atomic mass could remain, altering the structure of the core, nuclear reaction rates etc. The trouble is, that the vaporisation would occur whilst the contracting Sun was still fully convective, effectively mixing everything up throughout the Sun before nuclear reactions began.
Best Answer
The symmetry of the Sun has got very little to do with any symmetry in its formation.
The Sun has had plenty of time to reach an equilibrium between its self gravity and its internal pressure gradient. Any departure from symmetry would imply a difference in pressure in regions at a similar radius but different polar or azimuthal angles. The resultant pressure gradient would trigger fluid flows that would erase the asymmetry.
Possible sources of asymmetry in stars could include rapid rotation or the presence of a binary companion, both of which break the symmetry of the effective gravitational potential, even if the star were spherically symmetric. The Sun has neither of these (the centrifugal acceleration at the equator is only about 20 millionths of the surface gravity, and Jupiter is too small and far away to have an effect) and simply relaxes to an almost spherically symmetric configuration.
The relationship between oblateness/ellipticity and rotation rate is treated in some detail here for a uniform density, self-gravitating spheroid and the following analytic approximation is obtained for the ratio of equatorial to polar radius $$ \frac{r_e}{r_p} = \frac{1 + \epsilon/3}{1-2\epsilon/3}, $$ where $\epsilon$, the ellipticity is related to rotation and mass as $$\epsilon = \frac{5}{4}\frac{\Omega^2 a^3}{GM}$$ and $a$ is the mean radius, $\Omega$ the angular velocity.
Putting in numbers for the Sun (using the equatorial rotation period), I get $\epsilon=2.8\times10^{-5}$ and hence $r_e/r_p =1.000028$ or $r_e-r_p = \epsilon a = 19.5$ km. Thus this simple calculation gives the observed value to a small factor, but is obviously only an approximation because (a) the Sun does not have a uniform density and (b) rotates differentially with latitude in its outer envelope.
A final thought. The oblateness of a single star like the Sun depends on its rotation. You might ask, how typical is the (small) rotation rate of the Sun that leads to a very small oblateness? More rapidly rotating sun-like (and especially more massive) stars do exist; very young stars can rotate up to about 100 times faster than the Sun, leading to significant oblateness. However, Sun-like stars spin-down through a magnetised wind as they get older. The spin-down rate depends on the rotation rate and this means that single (or at least stars that are not in close, tidally locked binary systems) stars converge to a close-to-unique rotation-age relationship at ages beyond a billion years. Thus we expect (it remains to be proven, since stellar ages are hard to estimate) that all Sun-like stars with a similar age to the Sun should have similar rotation rates and similarly small oblateness.