Let $A = \sum_{a\in[K]} w_ax_ax_a^T$, where $w_a \in\mathbb{R}$, $x_a\in\mathbb{R}^d$, $K> 1$. Is there a way to write $$ A = \tilde{A_1}\otimes \tilde{A_2} $$ for some matrix $\tilde{A_1},\tilde{A_2}$?
Note that $A$ can be rewritten as $$ A = \sum_{a\in[K]} (\sqrt{w_a}x_a)\otimes (\sqrt{w_a}x_a)^\top, $$
and hence the question can be reformulated as finding matrix $\tilde{A_1}$ and $\tilde{A_2}$ such that $A = \sum_{a \in[K]} A_a \otimes A_a^\top = \tilde{A_1} \otimes\tilde{A_2}$, for $A_a = \sqrt{w_a}x_a$.
The answer is no. As a counterexample, take $$ C = \pmatrix{1&0\\0&0}, \quad D = \pmatrix{1&1\\1&1}. $$ Denote $$ A = C \otimes I + I \otimes D = \pmatrix{2 & 1 & 0 & 0\\1 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0& 1 & 1}. $$ There are no matrices $A_1,A_2$ such that $A_1 \otimes A_2 = M$ except in the trivial cases where either $A_1$ or $A_2$ have size $1 \times 1$. It is easy to see that $A_1,A_2$ cannot both be of size $2\times 2$, otherwise, the upper-left and lower-right $2 \times 2$ blocks would need to be multiples of each other.
We can express $A$ in the summation form, however, with $K = 3$ and $$ w_1 = 3/2, \quad w_2 = 1/2, \quad w_3 = 1,\\ x_1 = (1,1,0,0), \quad x_2 = (1,-1,0,0), \quad x_3 = (0,0,1,1). $$
Note that $$ \operatorname{rank}(A_1 \otimes A_2) = \operatorname{rank}(A_1) \operatorname{rank}(A_2). $$ If $A_1 \otimes A_2$ has size $4 \times 4$ and neither $A_1$ nor $A_2$ has size $4 \times 4$, then the ranks of $A_1,A_2$ must be either $1$ or $2$. Thus, it is impossible to have $\operatorname{rank}(A_1 \otimes A_2) = 3$. However, $\operatorname{rank}(A) = 3$.