A quick note: in each row and column of 4x4-cell squares, each number must appear in a multi-cell box exactly once. Proof left to reader.
In the bottom right square, the 6 needs to span both of the middle rows, meaning the 2x2 must be 6. Similarly, the 2 in this square must be in the upper right corner, since it can only be in the far right column; this fills the "only far column" boxes in this square, so the 1 is forced as well. With this 1 placed, we force the 1 in the lower left square to be in the upper left corner, and the 1 in the middle right square to be in the middle right box.
Using similar expanded Sudoku logic, the 3 in the middle left square must be in the lower right corner, since it has to cover both of the bottom two rows. Moreover, the 3 in the bottom right square must be in the upper left corner, since this is the only multi-cell block in the right column of squares that is not already barred from being a 3. This forces the 3 in the upper right square, and then the 3 in the upper left as well. The grid thus far:
Continuing in the right column of squares:
The 7 must be in the lower right corner of the lower right square, since it must cover the rightmost 2 columns, forcing the 7 in the bottom middle square. Now consider 4s: at least one of the multi-cell boxes in the right column of squares must be a 4. It cannot be either lower-right L, and it cannot be the upper left L in the middle right square, since it would block any 4 in the upper right square. So the middle 2x2 in the middle right square must be 4. We also see the 6 in this square must be in the middle left cell, forcing the 9 to be the upper left L, and also forces the 9 in the bottom right square. Continuing in the middle right square, we see the remaining boxes are 5 and 8. But the 8 cannot be in the single cell box, because then there would be no multi-cell box in the middle row of squares that could contain 8. This lets us finish the middle right and bottom right squares off. Progress so far:
Some additional deductions:
The only multi-cell box in the middle row of squares that can accommodate 1 is the lower right L of the middle box. All lower left Ls must contain each number once, so the upper left one must be 6, which lets us complete all the 6s. Additional Sudoku deductions let us fill more of the grid, in particular filling a 9 in the lower right L of the upper middle square. This leaves only the lower left 2x2 to accommodate a 9, and lets us finish all of the 9s and 1s. Additional deductions with standard Sudoku logic ensue, yielding:
4, 5, 7 and 8 are the remaining digits. We cannot absolutely place them in the upper row of squares, but we can use color to associate which squares must be the same:
Note particularly that blue must be one of 4 or 8, while either pink or purple could be 7. Looking in the first column, we see that 7 cannot be in the upper left of the middle left square, since that would leave no place for the 7 in the upper left square, and there is already a 7 in the lower right square. So pink must be 7. The grid fills in at this point.