You need to know the equation of state for the star's interior. Once you know this you can calculate the density variation with depth and the gravity inside the star.
Google for something like "star equation of state" to find lots of articles on the subject, but note that it's exceedingly complicated because there are so many factors at work. This is the sort of article you'll find: good luck reading it!
Note also that while we can use models to calculate equations of state, the results are only as good as the models. It's hard to know how good our models are when all we can see is the surface of the star.
After I commented on the question I started wondering what an observer inside a collapsing shell would experience.
If you construct a spherical shell then an observer inside it feels no gravity. This is true in Newtonian gravity, and is also true in General Relativity as a consequence of Birkhoff's theorem i.e. the metric inside the shell is the Minkowski metric.
In principle we can take the shell and compress it until it's external radius falls below the Schwarzschild radius $r = 2GM/c^2$, at which point the shell will start collapsing inwards and form a singularity in a finite time. In fact it's a very short time indeed. Calculating the lapsed time to fall from the horizon to the singularity of an existing black hole is a standard exercise in GR, and the result is:
$$ \tau \approx 6.57 \frac{M}{M_{Sun}} \mu s $$
That is, for a black hole of 10 solar masses the fall takes 65.7 microseconds! I would have to indulge in some head scratching to work out if the same time would be measured by an observer riding on the collapsing shell, but if the time isn't the same it will be of a similar order of magnitude. This means much of the question doesn't apply, since the shell cannot be stable long enough for the black hole to evaporate. However it leaves open the interesting question of what the observer inside the shell experiences.
Curious as it seems, Birkhoff's theorem implies the observer experiences absolutely nothing until the collapsing shell hits them and sweeps them, along with the shell, to an untimely end (a few microseconds later!).
Response to comment: time dilation
The infall time I calculated above is the proper time, that is the time measured by the freely falling observer on their wristwatch. You need to tread carefully when talking about time in relativity, but the proper time is usually easy to understand.
Re time dilation: again we need to be careful to define exactly what we mean. In the context of black holes we usually take an observer far from the black hole (strictly speaking at an infinite distance) as a reference and compare their clock to a clock near the black hole. By time dilation we mean that the observer at infinity sees the clock near the black hole running slowly.
A clock in a gravitational potential well runs slowly compared to the clock at infinity. This was discussed in the higher you go the slower is ageing (and also in Gravitational time dilation at the earth's center). It's important to understand that it's the potential that matters, not the gravitational acceleration, so even though the observer inside the shell feels no gravitational acceleration they are still time dilated compared to the observer at infinity.
Note that the time dilation relative to the observer at infinity goes to infinity at the event horizon, so it makes no sense to compare times inside the event horizon to anything outside.
Best Answer
It's because the value of the gravitational field at the center of a star is not the relevant quantity to describe gravitational collapse. The following argument is Newtonian.
Let's assume for simplicity that the star is a sphere with uniform density $\rho$. Consider a small portion of the mass $ m$ of the star that's not at its center but rather at a distance $r$ from its center. This portion feels a gravitational interaction towards the other mass in the star. It turns out, however, that all of the mass at distances greater than $r$ from the center will contribute no net force this portion. So we focus on the mass at distances less than $r$ away from the center. Using Newton's Law of Gravitation, one can show that the net result of this mass is to exert a force on $ m$ equal in magnitude to \begin{align} F = \frac{G( m)(\tfrac{4}{3}\pi r^3 \rho)}{r^2} = \frac{4}{3}G m\pi\rho r \end{align} and pointing toward the center of the star. It follows that unless there is another force on $m$ equal in magnitude to $F$ but pointing radially outward, the mass will be pulled towards the center of the star. This is basically what happens when stars exhaust their fuel; there no longer is sufficient outward pressure to counteract gravity, and the star collapses.