How can the tidal force due to a black hole tear apart the observer inside? I have heard that the tidal force of a black hole can tear apart an observer. But if the spacetime fabric is really stretched out near a black hole, isn't the observer also a part of it? Surely, he is not sitting on top of it! How could he distinguish himself being stretched, given everything around him also is getting stretched?
 A: If the observer were a part of the fabric, we would never have made the observations which confirm general relativity, of an expanding universe. Everything would be expanding and it would not be possible to see any differences. What happens after the inflation period is that the forces that bind galaxies and clusters of galaxies, (gravitational), electrons and nucleons into atoms, (electromagnetic), quarks and gluons into nuclei, (strong) are much stronger than the expansion of space and remain unchanged as spacetime expands. It is only in the inflation period that the equivalent expansion forces are stronger than the eventual four forces which create matter. 
The opposite happens when falling into a black hole, a point is reached where the space distortions are larger than the forces holding atoms and molecules together, destroying them.
A: Consider a star that approaches a black hole. Then to first order, we can consider that when the star passes close enough to the black hole such that the tidal force is comparable to the self gravity of the star, the tidal force can dominate and rip the star apart. This distance criterion can be described as:
$$ r_t = \left(\frac{M_h}{M_*}\right)^{1/3}R_*$$
Obviously there is also a dependence on the material of the star and the initial geometry. But as a basic though experiment, we can consider the star as a uniform sphere of gas particles that are purely gravitationally bound and still capture most of the physics.
Intuitively, the way to think about it is that since the force of gravity scales inversely to second power with radius, if we get close to a large source, the gravitational force difference between the front and back side of the star is large enough that the self gravity can no longer hold it together and it is ripped apart.
A: I like the question. I assume you have the free-falling elevator1 in mind: Its occupant perceives no acceleration and zero gravitation because there is really no difference: s/he is as much existing within space-time as anybody else and hence no view is favored, and no force is exerted in that frame of reference. 
So the question really is: Why should that be different in an inhomogeneous field? What can the poor black hole observer see that the poor falling elevator observer could not?
I assume the answer is that due to the inhomogeneity of the gravitational field, different parts of the observer experience different space-time geometries and hence want to "rest" in "different space-time places"; an outside observer would say the different parts have an urge to follow different trajectories — an urge which will eventually be granted down to the elementary particle level.

1 in a homogeneous gravitational field where thrown balls follow parabolas...
