The trace/Frobenius product is a convenient infix notation for the trace, i.e.
$$A:B = {\rm Tr}(A^TB)$$
The cyclic property of the trace allows terms in a Frobenius product to be rearranged in a myriad of ways, e.g
$$\eqalign{
A:BC &= BC:A &= A^T:(BC)^T \cr&= AB^T:C &= C^TA:B \,\,= I:A^TBC = etc. \cr
}$$
Consider a scalar function of two matrices
$$\lambda(X,Y) = AYA:X = A^TXA^T:Y$$
and calculate its differential.
$$\eqalign{
d\lambda &= AYA:dX + A^TXA^T:dY \cr
}$$
Now assume that these two matrices are function of a third matrix $$X=Y=Q^TQ$$
and we're asked to calculate the differential and gradient wrt $Q$.
$$\eqalign{
d\lambda
&= (AYA + A^TXA^T):(dQ^TQ+Q^TdQ) \cr
&= (AYA + A^TYA^T + A^TXA^T + AXA):Q^TdQ \cr
&= Q\Big(A(X+Y)A + A^T(X+Y)A^T\Big):dQ \cr
&= 2Q\Big(AQ^TQA + A^TQ^TQA^T\Big):dQ \cr
\frac{\partial\lambda}{\partial Q} &= 2Q(AQ^TQA + A^TQ^TQA^T) \cr
}$$
This is the gradient with respect to the full matrix $Q$.
To find the gradient with respect to the $k^{th}$ column, multiply by $e_k$ from the standard basis.
$$\eqalign{
q_k &= Qe_k \cr
\frac{\partial\lambda}{\partial q_k} &= \Big(\frac{\partial\lambda}{\partial Q}\Big)\,e_k \cr
&= 2Q\Big(AQ^TQA + A^TQ^TQA^T\Big)\,e_k \cr
}$$