Consider a black hole of some radius $R$. Now, suppose a cosmohiker (with her extremely powerful rockets and a glass bowl) enters the black hole. The mass of the cosmohiker along with her belongings is $m$. As they cross the horizon (according to a good choice of coordinates in which this happens), the radius of the black hole increases a little bit along with its entropy. But then onwards, the entropy of the black hole remains constant. But if the regular laws of thermodynamics were valid in the interior (not far beyond the horizon) then the entropy of the black hole should have increased as the cosmohiker breaks her glass bowl. But to my understanding, this doesn't happen according to the area law. The black hole entropy is strictly a function of its area only and that is not dependent on what is going on with the stuff that went inside the black hole - it is dependent on their total mass only. So, do the laws of thermodynamics breakdown immediately inside the event horizon?
Once they cross the horizon pretty quickly the horizon grows a bit (actually even as it gets very close to the horizon, a little horizon bubble forms). The horizon grows as the bubble gets absorbed, till it absorbs it all, and so the radius and the horizon area grows, and the entropy. It happens very quickly -- the final merger of the binary BH's in 2015 took less than a second, with the final ringdown even less.
The dynamic process as it is merging can be done only approximately and through a numerical simulation. For small masses being absorbed you can do a perturbation simulation, always being careful with the coordinate singularity at the horizon. BH thermodynamics describes the end states.
The entropy grows proportionately to the area, and the area proprortionately to the the mass for a Schwarzschild BH. For a more general BH, the 1st law of BH thermodynamics the angular momentum and charge also contribut to the mass (i.e., BH energy). And the 2nd law says entropy increases as the horizon is crossed and equilibrium is achieved. After that there is no more entropy or area change, or total mass change (until something else happens).
See BH thermodynamics at https://en.wikipedia.org/wiki/Black_hole_thermodynamics
See the 2015 binary BH merger simulation at https://en.wikipedia.org/wiki/Binary_black_hole
This is actually a very good question!
Let's consider the Einstein equations in all of their glory:
(Note that I am using units where $c=8\pi G=1$, so that the Schwarzschild radius is given by $M/4\pi$). Black holes are solutions to the vacuum Einstein equations, namely
Now, one the fundamental feature of a black hole solution is that it is eternal. That is, its metric does not depend on time. A black hole solution cannot form in a finite amount of time. Another fundamental property of black holes is that they are completely described by three numbers: their mass, their angular momentum, and their electric charge. This is known as the no hair theorem.
Now, if we include quantum effects for a black hole solution whose event horizon area is $A$, then the entropy is given by
(Someone should check that my factors are right.) If I set $G=1$ instead then I would have $S=A/4$, which is the usually quoted form. The form of this entropy is actually a pretty good example of the no hair theorem: the area only depends on the mass, angular momentum, and electric charge, and the entropy only depending on these things implies that there is no more complicated structure to the black hole.
Okay, so I've covered some of the fundamentals. Now, in your example, you have a complicated system, namely a cosmohiker with a glass bowl. This system has a highly nontrivial (and, most importantly, nonzero) stress energy tensor. Thus, the metric will no longer satisfy the vacuum Einstein equations. That is, the giant celestial object that the cosmohiker is approaching can is no longer a pure black hole: it is perturbed by her presence near it! Thus, simply due to the presence of an object with nonzero energy momentum tensor, the black hole itself gets "hairy" (that is, it becomes more complicated than a pure black hole). Since the entropy formula above only applies to pure black holes, it no longer applies.
This is, in essence, why your intuition breaks down here. In the presence of the cosmohiker, the entropy becomes more complicated and no longer is just dependent on the area of the black hole itself. However, long after the cosmohiker falls to the singularity and dies, the solution asymptotically approaches that of a pure black hole with an increased area, and the entropy will eventually approach the form above. But within the short times in which the cosmohiker does her antics (breaking bowls and whatnot -- silly cosmohiker), the entropy does not follow this area law exactly.
Thanks for a great question. I hope this helped!