Just add the indices. In the formulae above, the doubly repeated indices are automatically summed over, using the Einstein sum rule. Note that ${\rm Tr} M = M_{ii}$ and $(MN)_{ij} = M_{ik}N_{kj}$.
The first derivative of the trace is $ \frac{\partial A_{ab}B_{bc}A^T_{ca} }{\partial A_{ij}} = \frac{\partial A_{ab}B_{bc}A_{ac} }{\partial A_{ij}} =$ $= \delta_{ai}\delta_{bj} B_{bc} A_{ac} + A_{ab} B_{bc} \delta_{ai}\delta_{cj} = B_{jc}A_{ic} + A_{ib}B_{bj} = (AB^T)_{ij} + (AB)_{ij} $ so by removing the indices $ij$ again, you see that the "matrix derivative" is $AB^T+AB$. Similarly, $ \frac{\partial A^T_{ab}B_{bc}A_{ca} }{\partial A_{ij}} = \frac{\partial A_{ba}B_{bc}A_{ca} }{\partial A_{ij}} = \delta_{bi}\delta_{aj} B_{bc} A_{ca} + A_{ba} B_{bc} \delta_{ci}\delta_{aj} =$ $ = B_{ic}A_{cj} + A_{bj}B_{bi} = (BA)_{ij} + (B^T A)_{ij} $ so the second result is $BA+B^TA$.