Why are the t2 orbitals above the e orbitals in a tetrahedral complex using the crystal field theory?
Take into account that $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ (or $\mathrm{e}$ and $\mathrm{t_2}$) are symmetry terms, and they make sense in the context of a certain symmetry. In other words, they do not describe something that is implicit to the orbitals, but, rather, how are orbitals affected by an environment with a certain symmetry.

In crystal field theory, we look only at the electrostatic effect of a certain distribution of charges around the nucleus on the orbitals; the discussion is primarily qualitative and geometric in nature.

In an environment with spheric symmetry (such as an isolated atom), all 5 d-orbitals are degenerate; since electrostatics are the same in any radial direction, and the set of 5 d-orbitals are only distinguished by their angular part (how are they distributed in the solid angle around the nucleus), they are equivalent. [Incidentally, this would be true not only for an isolated atom but also for an atom surrounded by a spherically symmetric charge distribution, such as a uniform charge distribution a certain distance away from the nucleus; however, I can't think of a real system that would have exact spherical symmetry.]

As soon as you introduce point charges around the nucleus, however, symmetry will break down. All 5 d-orbitals will no longer interact equally with the point charge; some orbitals will be closer to the point charge, some will be further away; and, most importantly from the point of view of symmetry, perhaps several orbitals will interact equally with the point charges.

This may be easier to visualize with the three $\mathrm p_x$, $\mathrm p_y$, $\mathrm p_z$ orbitals: if there are two point charges along the $z$ axis, it is clear that the $\mathrm p_z$ will interact more strongly with them than the $\mathrm p_x$ and $\mathrm p_y$ orbitals, but, crucially, $\mathrm p_x$ and $\mathrm p_y$ will interact exactly the same way with the charges; their interactions with the point charges are symmetry-equivalent.

In the extreme case where the distribution of point charges around the atom has no symmetry, all five d-orbitals would in general have different energies, because their interactions with the charges would be different; but if our distribution of charges is symmetric (as happens in both the octahedral and the tetrahedral case), we need to look at how the orbitals behave under the symmetry.

If you do that (and Jan's excellent answer explains what happens with the Cartesian axes under both symmetries), you will see that in the octahedral case two orbitals behave the same way (and are destabilised to the same extent), while another set of three orbitals behave equally (and are stabilised to the same extent). Looking further at their symmetry properties, these sets of two and three orbitals behave in the ways we describe with $\mathrm{e_g}$ and $\mathrm{t_{2g}}$, respectively. In the tetrahedral case, per contrast, we find that two orbitals are equally stabilised while three orbitals are equally destabilised; the symmetry label that fits their behaviour is $\mathrm e$ and $\mathrm t_2$. [In both cases, $\mathrm e$/$\mathrm{e_g}$ is a symmetry label that corresponds to two elements, and $\mathrm t_2$/$\mathrm{t_{2g}}$ corresponds to three.] These "blocs" of orbitals are degenerate (i.e. these two or three orbitals all have the same energy because they interact similarly with the point charges in the environment), while the energy of these blocs is not the same - that's what we call "splitting".

Note that the temptation to think that there are two "$\mathrm e$" d-orbitals and three "$\mathrm t_2$" d-orbitals in an atom is to be avoided: there is nothing intrinsic in the orbitals that produces this particular splitting into two blocs; rather, it's a product of the symmetry of the environment. The same 5 orbitals, if put in a square planar environment, do not form blocs of 2 and 3 orbitals, but 2, 1, 1 and 1; in a trigonal bipyramidal environment, they split into 2, 2, and 1; and so on.

So it just happens that two orbitals are stabilised in a tetrahedral environment (and thus are labelled $\mathrm e$) and three are destabilised (thus labelled $\mathrm t_2$), while it's the opposite in an octahedral environment.

[As a thought exercise: in crystal field theory, we assume that we are dealing with a central cation surrounded by anions, because that's the problem for which CFT was developed. But, what would happen if the point charges around the atoms were positive instead of negative? You'll find that you end up with stabilised $\mathrm{e_g}$ (octahedral) and $\mathrm t_2$ (tetrahedral) and destabilised $\mathrm{t_{2g}}$ (octahedral) and $\mathrm e$ (tetrahedral), exactly the opposite splitting but the same symmetry behaviour.]

There are rigorous mathematical ways to deduce the symmetry labels of d-orbitals (or any orbital, for that matter) under an environment of a certain symmetry, as provided by point group theory, which involve symmetry tables and so on. Most likely, your textbook will include a discussion of the derivation for at least the octahedral case.