If you're not up to speed with general relativity this is going to be hard to explain, but I'll give it a go. The more determined reader may want to look at this PDF (just under 1MB in size) that describes the collapse in a rigorous way.
A couple of points to make before we start: you're being vague about the distinction between the singularity and the event horizon. The singularity is the point at the centre of the black hole where the curvature becomes infinite. The event horizon is the spherical surface that marks the radial distance below which light cannot escape. As you'll see, these form at different times.
The other point is that to make the calculation possible at all we have to use a simplified model. Specifically we assume the collapsing body is homogeneous (actually I see you anticipated that in your answer) and is made up of dust. In general relativity the term dust has a specific meaning - it means matter that is non-interacting (apart from gravity) and has zero pressure. This is obviously very different from the plasma found in real stars.
With the above simplifications the collapse is described by the Oppenheimer-Snyder model, and it turns out that the size of the collapsing object is described by the same equation that describes the collapse of a closed universe. This equation is called the FLRW metric, and it gives a function called the scale factor, $a(t)$, that describes the size of the ball of dust. For a closed universe the scale factor looks something like:
(image from this PDF)
A closed universe starts with a Big Bang, expands to a maximum size then recollapses in a Big Crunch. It's the recollapse, i.e. the right hand side of the graph above, that describes the collapse of the ball of dust.
The radius of the ball is proportional to $a(t)$, so the radius falls in the same way as $a(t)$ does, and the singularity forms when $a(t) = 0$ i.e. when all the matter in the ball has collapsed to zero size.
As always in GR, we need to be very careful to define what we mean by time. In the graph above the time plotted on the horizontal axis is comoving or proper time. This is the time measured by an observer inside the ball and stationary with respect to the grains of dust around them. It is not the same as the time measured by an observer outside the ball, as we'll see in a bit.
Finally, we should note that the singularity forms at the same time for every comoving observer inside the ball of dust. This is because the ball shrinks in a homogeneous way so the density is the same everywhere inside the ball. The singularity forms when the density rises to infinity (i.e. the ball radius goes to zero), and this happens everywhere inside the ball at the same time.
OK, that describes the formation of the singularity, but what about the event horizon. To find the event horizon we look at the behaviour of outgoing light rays as a function of distance from the centre of the ball. The details are somewhat technical, but when we find a radius inside which the light can't escape that's the position of the event horizon. The details are described in the paper by Luciano Rezzolla that I linked above, and glossing over the gory details the result is:
This shows time on the vertical axis (Once again this is comoving/proper time as discussed above) and the radius of the ball of dust on the horizontal axis. So as time passes we move upwards on the graph and the radius decreases.
It's obviously harder for light to escape from the centre of the ball than from the surface, so the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface when the radius of the ball has decreased to:
$$ r = \frac{2GM}{c^2} $$
This distance is called the Schwarzschild radius and it's the event horizon radius for a stationary black hole of mass $M$. So at this moment the ball of dust now looks like a black hole and we can no longer see what's inside it.
However note that when the event horizon reaches the Schwarzschild radius the collapse hasn't finished and the singularity hasn't formed. It takes a bit longer for the ball to finish contracting and the singularity to form. The singularity only forms when the red line meets the vertical axis.
Finally, one last note on time.
Throughtout all the above the time I've used is proper time, $\tau$, but you and I watching the collapse from outside measure Schwarzschild coordinate time, $t$, and the two are not the same. In particular our time $t$ goes to infinity as the ball radius approaches the Schwarzschild radius $r = 2GM/c^2$. For us the part of the diagram above this point simply doesn't exist because it lies at times greater than infinity. So we never actually see the event horizon form. I won't go into this any further here because it's been discussed to death in previous questions on this site. However you might be interested to note this is one of the reasons for Stephen Hawking's claim that event horizons never form.
Time and space don't swap places inside a black hole.
In order to describe what happens in spacetime we have to attach labels to spacetime points, and these labels are our coordinate system. That is, we choose some coordinates $t$, $r$, $\theta$ and $\phi$ then we can label points in spacetime by their coordinates $(t, r, \theta, \phi)$.
But there are lots of ways to form a coordinate system. The usual one for describing the exterior of black holes are called the Schwarzschild coordinates and these correspond to the physical measurements made by an observer an infinite distance from the black hole. So the Schwarzschild $t$ coordinate is the time measured on the Schwarzschild observer's clock. However there are lots of other ways to make a coordinate system. Gullstrand-Painlevé coordinates are superficially similar but now $t$ is the time measured on a clock held by an observer falling freely into the black hole. Or Kruskal-Szekeres coordinates use abstract coordinates that don't correspond to anything measured by an observer.
The point of all this is that the coordinates are not spacetime - they are just labels we attach to spacetime. What happens inside an event horizon is not that time and space swap places but rather that the labels we call Schwarzschild coordinates behave oddly inside a black hole. This is an important distinction. If we use the Kruskal-Szekeres coordinates instead then we have a spacelike coordinate $u$ and a timelike coordinate $v$ and these do not swap places inside the black hole.
It is fairly easy to see why the Schwarzschild coordinates are not a good description for the interior of a black hole. If we drop something into a black hole and start timing its fall we discover that the object takes an infinite time to even reach the event horizon. That is, no matter how long we time for the falling object never reaches the horizon. In fact for Schwarzschild observers black holes don't exist because any black hole will take an infinite time to form. So by using Schwarzschild coordinates to label the interior of a black hole we are labelling something that doesn't exist. Is it then any wonder that the behaviour of our labels, our coordinate system, inside the black hole is bizarre?
Where things get confusing is that even though in the Schwarzschild coordinates black holes can never form this does not mean they don't exist. For example if you're standing on the surface of a collapsing star then in your rest frame not only does the black hole form in a finite time, but you will fall through the event horizon and to your fatal encounter with the singularity in a finite time. The Schwarzschild $t$ coordinate only covers your trajectory up to the formation of the horizon, but your trajectory carries on past the point where the Schwarzschild $t$ coordinate reaches infinity. But once again let me emphasise that this is just a failing of the Schwarzschild coordinate system. Time carries on for you as normal - it's just just that the Schwarzschild coordinates are not a good way to label time in those circumstances.
Best Answer
Indeed you made one mistake: the infalling observer does not see the outside universe "speed up". Look at what happens in a space-time diagram. At the spacetime point where your astronaut passes the horizon, he can only see what's in his past light cone, and that's the universe at early times only. It's the signals that he sends back (or tries to) that reach the outsine world only at infinite times.
Thus, the observing astronaut sees his black hole as it is long before any evaporation sets in, so his black hole is still there. Now leaving aside some other quantum issues, where opinions aren't completely settled and perhaps even our presently used language could be inapropriate, the observer just continues on, and in a finite amount of time, very quickly unless the black hole were more than millions of times heavier than the sun, he is killed by the central singularity.
In a black hole with high angular momentum (Kerr black hole), the singularity takes the form of a ring along the equator, and the astronaut might try to sail past it safely, and he would be able to enter into a strange new universe where he may or may not leave a negative mass black hole behind him, were it not that debris from other objects that fall in later will kill him before that happens, and while trying to pass a second horizon, he will be killed because that second horizon is unstable.