Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?
The answer is, as @orthocresol mentioned, due to two parts.
Initially, if we are assuming the ligands are point charges (crystal field theory) then in octahedral complexes, the ligands are more closely associated with the d orbitals, as shown in this diagram: 
But, in tetrahedral complexes, the ligands are not so closely associated with the d orbitals, as shown in this diagram: 
By associating the d orbitals seen in the first diagram, with the tetrahedral point charges in the second diagram, you can see how close the point charges are to the d orbitals in the octahedral case compared to tetrahedral.
Also as otho mentioned, octahedral has 6 ligands interacting with 5 d orbitals, whereas tetrahedral has only 4 ligands interacting with 5 d orbitals. Octahedral has the more negative charge due to more electrons, therefore more repulsion.