Result: $\operatorname{vec}(A \otimes B) = [I_m \otimes K^{(q,m)} \otimes I_q][\operatorname{vec}(A)\otimes \operatorname{vec}(B)]$, where $I_k$ denotes the identity matrix of size $k$ and $K^{(q,m)}$ denotes a commutation matrix.
Proof: It suffices to consider the case that $A = uv^T$ and $B = xy^T$ (for column vectors $u,v,x,y$). We have
$$
\operatorname{vec}(A \otimes B) = \operatorname{vec}[(uv^T) \otimes (xy^T)] =
\operatorname{vec}[(u \otimes x)(v \otimes y)^T]
= v \otimes y \otimes u \otimes x.
$$
It now suffices to use the fact that $K^{(q,m)}(u \otimes y) = y \otimes u$ to show that the matrix $P = I_m \otimes K^{(q,m)} \otimes I_q$ satisfies
$$
P(\operatorname{vec}(A) \otimes \operatorname{vec}(B)) =
P(v \otimes u \otimes y \otimes x)=
v \otimes y \otimes u \otimes x = \operatorname{vec}(A \otimes B).
$$
One approach is to note that $\mathcal S$ can be decomposed as the sum of simple tensors,
$$
\mathcal S =\sum_{j=1}^r \bigotimes_{i=1}^d S_{ij} = \sum_{j=1}^r S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j},
$$
where $S_{ij} \in \Bbb R^{n_i}$ for all $j$ and $\bar \otimes$ denotes the "tensor product". Let $\phi_M^{(k)}$ denote the linear map associated with the modal product taken here. That is, $\phi_{M}^{(k)}$ is defined so that $\phi_M^{(k)}(\mathcal S) = \mathcal A$.
Note/show the following facts:
For all $j$,
$$
\phi_M^{(k)}[S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j}] = \\
S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes (MS_{k,j})\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j}
$$
For all $j$,
$$
\operatorname{vec}[S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j}] = \\
S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes S_{k,j} \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}
$$
Denoting $ \hat M_k = I_{n_{k+1}\ldots n_d}\otimes M\otimes I_{n_1\ldots n_{k-1}}$, we have
$$
\hat M_k(S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes S_{k,j} \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}) = \\
S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes (MS_{k,j}) \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}
$$
From there, the proof is straightforward:
\begin{align}
\operatorname{vec}(\mathcal A) &= \operatorname{vec}
\left[\phi_M^{(k)}(\mathcal S)\right]
\\ & =
\operatorname{vec}
\left[\phi_M^{(k)}\left[\sum_{j=1}^r S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j} \right]\right]
\\ & =
\operatorname{vec}
\left[\sum_{j=1}^r \phi_M^{(k)}\left[ S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j} \right]\right]
\\ & =
\operatorname{vec}
\left[\sum_{j=1}^r S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes (MS_{k,j})\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j} \right]
\\ & =
\sum_{j=1}^r \operatorname{vec}
\left[ S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes (MS_{k,j})\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j} \right]
\\ & =
\sum_{j=1}^r S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes (MS_{k,j}) \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}
\\ & =
\sum_{j=1}^r \hat M\left[S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes S_{k,j} \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}\right]
\\ & =
\hat M\left[\sum_{j=1}^r S_{d,j} \otimes \cdots \otimes S_{k+1,j} \otimes S_{k,j} \otimes S_{k-1,j} \otimes \cdots \otimes S_{1,j}\right]
\\ & =
\hat M\left[\sum_{j=1}^r \operatorname{vec}\left[S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j}\right]\right]
\\ & =
\hat M \operatorname{vec}\sum_{j=1}^r \left[S_{1,j} \bar \otimes \cdots \bar \otimes S_{k-1,j} \bar \otimes S_{k,j}\bar \otimes S_{k+1,j} \bar \otimes \cdots \bar \otimes S_{d,j}\right]
= \hat M \operatorname{vec}(\mathcal S),
\end{align}
which is what we wanted.
We could also save ourselves a lot of notational headache if we prove it for $k = 1$, then generalize that proof as follows. Let $\tau$ denote the "transpose" map over $\Bbb R^{n_1 \times \cdots \times n_d}$ that switches dimensions $1$ and $k$ of $\mathcal S$. If I've understood your terminology correctly, then I believe this means we're swapping the $1$st and $k$th "modes". Similarly, let $\sigma$ denote the map that does the same thing for $\mathcal A$ (noting that $\mathcal S$ and $\mathcal A$ have different shapes).
Note that there is a permutation matrix $P_\tau$ such that
$$
\operatorname{vec}(\tau(\mathcal S)) = P \operatorname{vec}(\mathcal S)
$$
holds for all $\mathcal S \in \Bbb R^{n_1 \times \cdots \times n_d}$, and similarly there is a corresponding $P_\sigma$. We can write
\begin{align}
\operatorname{vec}(\phi_M^{(k)} \mathcal S) & =
\operatorname{vec}(\sigma^{-1}(\phi_M^{(1)} \tau(\mathcal S)))
\\ & =
P_{\sigma}^{-1}\operatorname{vec}(\phi_M^{(1)} \tau(\mathcal S))
\\ & =
P_{\sigma}^{T}\hat M_{1} \operatorname{vec}(\tau(\mathcal S))
=
P_{\sigma}^{T} \hat M_1 P_{\tau} \operatorname{vec}(\mathcal S).
\end{align}
So, it suffices to show that $P_{\sigma}^{T} \hat M_1 P_{\tau} = \hat M_k$. To that end, it suffices to show that $v^\top P_{\sigma}^{T} \hat M_1 P_{\tau} w = v^\top \hat M_k w$ for all appropriately sized column-vectors $v,w$. Moreover, it suffices to show that this holds for vectors of the form $v = v_1 \otimes v_2 \otimes v_3$ and $w = w_1 \otimes w_2 \otimes w_3$ where $v_1,w_1 \in \Bbb R^{n_1 \cdots n_{k-1}}$, $v_2 \in \Bbb R^{m_k}$, $w_2 \in \Bbb R^{n_k}$, and $v_3,w_3 \in \Bbb R^{n_{k+1} \cdots n_d}$.
Best Answer
$\def\v{\operatorname{vec}}\def\o{\otimes}\def\B{\,\color{red}{B}\,}$By reshaping $\v(B)$ we can create a matrix $\B$
Consider the case for $m=4$ $$\eqalign{ \v\Big((A_1) \B (A_4^T\o A_3^T\o A_2^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ \v\Big((A_2\o A_1) \B(A_4^T\o A_3^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ \v\Big((A_3\o A_2\o A_1) \B(A_4^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ }$$ Extrapolating, if the number of $A$ matrices is $m$, then there are $(m-1)$ locations where the $\color{red}{B}$ matrix can be placed, which will yield the same result upon vectorization. The $A$ matrices are cyclically permuted to accommodate the shifting $\color{red}{B}$ matrix.
Another approach is calculate the Kronecker factorization of the vector into constituent vectors whose dimensions are compatible with the $A$ matrices $$\eqalign{ &\v(B) = b_4\o b_3\o b_2\o b_1 \\ &(A_4\o A_3\o A_2\o A_1)\v(B) = (A_4b_4\o A_3b_3\o A_2b_2\o A_1b_1) \\ }$$ Although it may not be possible to find a monomial factor, it is always possible to find a sum of such factors, e.g. $$\eqalign{ \v(B) &= \sum_{k=1}^K b_{4k}\o b_{3k}\o b_{2k}\o b_{1k} \\ }$$