Let $$y = Ax$$ $$z = y^Ty$$.
Per the chain rule, $$\frac{\partial{z}}{\partial{A}} = \frac{\partial{y}}{\partial{A}} \frac{\partial{z}}{\partial{y}} $$
Since,
$$\frac{\partial{z}}{\partial{A}} = \frac{\partial}{\partial{A}}(x^TA^TAx) = 2yx^T$$
$$\frac{\partial{y}}{\partial{A}} = \frac{\partial}{\partial{A}}(Ax) = x^T \otimes \Bbb{I}$$
$$\frac{\partial{z}}{\partial{y}} = \frac{\partial}{\partial{y}}(y^Ty) = 2y$$
It means that,
$$2yx^T = (x^T \otimes \Bbb{I}) \text { mystery product operator } 2y$$
What is the mystery product operator ?
Using index notation it's easier to see what's going on $$\eqalign{ \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\LR#1{\left(#1\right)} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} y_i &= A_{ip}\,x_p \\ \grad{y_i}{A_{jk}} &= \grad{A_{ip}}{A_{jk}}\;x_p \\ &= \delta_{ij}\delta_{pk}\,x_p \\ &= \delta_{ij}\,x_k }$$ where $x_k$ is the $k^{th}$ component of the $x$ vector and $\delta_{ij}$ is the $(i,j)$ component of the identity matrix. The resulting quantity is a third-order tensor.
So you don't need the Kronecker product $(\otimes),\,$ but rather the dyadic (aka tensor) product $(\star)$ $$ \frac{\partial y}{\partial A} = I\star x $$ Although some authors (especially in certain engineering disciplines) use the $(\otimes)$ symbol to denote the tensor product.
The chain rule uses an ordinary dot product $(\cdot)$ $$\eqalign{ \grad{z}{A} &= \gradLR{z}{y}\cdot \gradLR{y}{A} \\ &= \LR{2y}\cdot \LR{I\star x} \\ &= \LR{2y\cdot I}\star x \\ &= {2y\star x} \\ &= {2yx^T} \\ }$$