According to Susskind a bit of information crossing the event horizon of a black hole instantaneously encounters the singularity. Also, time appears to gradually slow down for an object approaching the horizon. Is this tantamount to saying that the curvature of spacetime approaches and becomes vertical exactly at the horizon? Or could you say that the curvature just becomes close enough to vertical at the horizon to trap light/information? If the latter is true then there is finite spacetime inside the horizon. This spacetime could not be observed by an observer outside the black hole. Again, referencing Susskind, black hole complementarity suggests that an observer passing into the horizon would not experience anything and that the observer would hit the singularity. Yes, but not instantaneously that would be paradoxical to experiencing nothing. The question then is when will the observer hit the singularity. And furthermore, could the space time inside the horizon be very large to the internal observer when you consider a variable transformation for the ratio of the speed of the clocks for the internal and external observers?
The general understanding is that inside a black hole space and and time change their places, see for example explanation here: https://www.einstein-online.info/en/spotlight/changing_places/ . If I understand it right, one should correctly speak about distance and not time for reaching the singularity. Another interesting and unconventional explanation I have heard is that if you crossed (without noticing) by car the event horizon on your Monday, then no matter where you go, how you drive, cross and across, back and forth, it comes Friday and you cease to exist in the singularity. It came me in mind that it is the best description of live, too. You are born without noticing, do a lot of things, driving, too, and on some day you cease to exist. Nice, isn't it?
A more enlightening explanation I have heard from Gerard t'Hooft is:
... . "An exact solution helps: consider a black hole formed by matter that goes in by the speed of light. Doesn't change the physics very much but makes it easy to understand. If all particles (basically without rest mass) would enter in a spherically symmetric mode then you can write the solution exactly. One finds that the horizon already opens up at a space-time point at the center (but no singularity there or anywhere else). As soon as the matter passed the horizon the outside world is in the Schwarzschild metric. Now you have to understand that in the inside region, surrounded by the horizon, space and time interchange roles. What you thought is space (such as the r coordinate) is actually time and what you thought to be time (the t coordinate) is actually space. The singularity is at r=0 but that is actually in the future. Not only that, it is, in a sense, the infinite future because outside observers will never see anything that has passed the horizon. For the outside observer, that never happens. For a black hole formed by matter, there is no past singularity. For quantum mechanics however, everything has to be reformulated. Singularities disappear or become physically immaterial. There are many more such things that people fail to understand, while it isn't difficult.
G. 't Hooft"
I hope, I could help.
An observer from outside the black hole watching someone travel towards the black hole will begin to see them slow down to an almost complete stop, and then they will slowly see them fading away since the photons that were being bounced off of the person falling into the black hole going towards the observer, cannot escape the immense pull of the gravitational field. Also, I don't think it would be accurate to say that the curvature becomes 'vertical' since depending on your model of the black hole, the geometry pass the horizon is definable. So your second response to your own question would be true, ''information passing the horizon cannot leave [unless via black hole evaporation which is a whole other issue].''
For example, the Schwarzschild metric omits a horizon from $f(r) = (1-2GM/r)\rightarrow r = 2GM$. But, you can get around this horizon and have smooth geometry past the horizon by switching to Kruskal-Szekeres coordinates. However, the physical singularity still remains, which is where curvature blows up ($Riem(g)\rightarrow \infty$).
With regards to switch space and temporal coordinates, yes this happens past the singularity if you assume continuity past the singularity. This is best represented by Penrose diagrams.
And finally, for the question of, ''when will the observer hit the singularity?'' For that, the person falling into the black hole will not instantaneously hit the singularity, but in a finite amount of time, they will eventually reach the singularity.
Hope that covers most of your question!
The question of when a bit of information reaches the singularity may not be the right question. How long it takes to get a bit of information out via Hawking radiation may be more to the point. Susskind says it takes about 10^66 years. So, the black hole does not function simply as a source of information for timeframes on the order of the age of the universe. Logically you would expect that massive black holes in the center of galaxies have heretofore acted as entropy sinks as one explanation why the observable universe seems to favor low entropy permutations. When two black holes collide, they merge to form a new single black hole. Assume the information for the newly formed black hole has information distributed across its event horizon in plank size elements and characterized as captured strings. One could speculate then than when a black hole forms from a collapsing star that its horizon merges with our universe’s surface and increases the information at the source of the holographic projection. This leads to an “it’s turtles all the way down” scenario but it preserves the second law in less than almost infinite time. There is also a symmetry here where our observable universe gains information and the black hole gains entropy. For an observer inside the black hole, I guess the opposite would be true.