The collapse of a real star is fearsomely complicated. We have no analytic solution of the Einstein equation for it so we have to simulate it numerically on a very large computer. However there is an idealised solution called the Oppenheimer-Snyder metric that captures the main features.
Your neutron star is supported by the neutron degeneracy pressure i.e. the Pauli exchange repulsion between neutrons. When you add the last (500kg) straw to the camel's back the compression becomes too great for the Pauli repulsion to resist. The star starts to collapse, and the event horizon forms first at the centre and then grows outwards towards the surface of the star. The horizon passes through the surface when the surface of the star has collapsed inwards to a distance equal to the Schwarzschild radius:
$$ r_s = \frac{2GM}{c^2} $$
The rate at which the horizon moves outwards depends on how fast the star collapses, and in real stars is likely to be a goodly fraction of the speed of light.
So if you place your 500kg object on the star to start the collapse you would have a few microseconds to pull it away again before the event horizon reaches the surface of the star. Note that pulling the object away will not stop the star collapsing. Once you have triggered the collapse it will continue until the star is consumed whether you pull the object away or not.
What you and the object would see is surprisingly complicated. For any observer outside the horizon it takes an infinite time for the horizon to form. That is, for an external observer the horizon never forms. However you would see an apparent horizon form in a few microseconds and this would be so similar to a real horizon that you could not tell the difference.
If you remain on the surface of the star as it falls inwards, so you fall inwards with it, you would see the horizon remaining just below you. As you fall in the horizon appears to retreat before you. You would only reach the horizon when you hit the singularity at the centre of the black hole.
