To determine the incorrect statement among the given options for the isoelectronic series $VO_4^{3-}$, $CrO_4^{2-}$, and $MnO_4^-$, we need to evaluate each statement based on the known properties of these species and their charge transfer (CT) transitions.

1. The statement "CT transitions are attributed to excitations of electrons from ligand (σ) to metal (e)" is generally true for charge transfer transitions. In these species, oxygen typically acts as a ligand, and the electron excitation occurs from the oxygen p-orbitals (ligand) to the metal's d-orbitals. Consequently, this statement is correct.
2. Next, consider the statement: "MnO_4^- exhibits charge transfer at the shortest wavelength among the three." In charge transfer transitions, a higher nuclear charge (or effective nuclear charge) typically leads to absorption at shorter wavelengths (higher energy). Among the three, MnO_4^- is known to have a higher nuclear charge compared to CrO_4^{2-} and VO_4^{3-}, which means it should indeed have the shortest wavelength transitions. However, the question asks for the incorrect statement, so let's explore further.
3. Consider the statement: "The wavelengths of transitions increase in the order VO_4^{3-} < CrO_4^{2-} < MnO_4^-." This implies that MnO_4^- would have the longest wavelength, which is inconsistent with the known behavior that MnO_4^- should have the shortest wavelength due to its high nuclear charge. Thus, this statement is incorrect.
4. The statement "The charge on the metal nucleus increases in the order VO_4^{3-} < CrO_4^{2-} < MnO_4^-" is correct. The charge on the metal indeed increases from Vanadium (+5), Chromium (+6), to Manganese (+7) in the given order.

Based on this analysis, the incorrect statement is: "MnO_4^- exhibits charge transfer at shortest wavelength among the three." Instead, it should exhibit the charge transfer at the shortest wavelength due to its highest nuclear charge leading to highest energy transition.