Christina sent us her solution:

A knight can't make a tour on a $2\times n$ board, for any $n$, because it must go into and out of a corner square, and it can't do this without going back on itself.

On the $3\times 4$ grid, we must use a path from the loop JAGIBHJ and a path from the loop KDFLCEK. But they only link up between J and C, and between B and K. So the path must start at a neighbour of J, B, K or C, follow round that loop, switch to the other loop and follow round that. Obviously the path can go round the loop in either direction. So there are $16$ possible tours:

HJAGIBKDFLCE

HJAGIBKECLFD

HBIGAJCEKDFL

HBIGAJCLFDKE

AGIBHJCEKDFL

AGIBHJCLFDKE

IGAJHBKECLFD

IGAJHBKDFLCE

ECLFDKBIGAJH

ECLFDKBHJAGI

EKDFLCJHBIGA

EKDFLCJAGIBH

DFLCEKBIGAJH

DFLCEKBHJAGI

LFDKECJAGIBH

LFDKECJHBIGA

Since it's not possible to get from the finish directly back to the start in any of these tours, there is no circuit.