Tensor (outer) product notation Consider two vectors (i.e. first-order tensors) $\boldsymbol{a}$ and $\boldsymbol{b}$ which can be expressed in index notation as $a_{i}\,\boldsymbol{e}_{i}$ and $b_{i}\,\boldsymbol{e}_{i}$ respectively.
These vectors have a scalar product given by
\begin{equation}
\boldsymbol{a}\cdot\boldsymbol{b}=a_{i}b_{i} \qquad\boldsymbol{a},\,\boldsymbol{b}\in\mathcal{R}^{3}\,,
\end{equation}
and an outer product, denoted by $\otimes$, that yields a second-order tensor $\boldsymbol{C}$ given by
\begin{align}
\boldsymbol{C}&=\boldsymbol{a}\otimes\boldsymbol{b}\\
&=a_{i}b_{j}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j} \\ 
&= C_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j} \qquad \boldsymbol{C}\in\mathcal{R}^{3}\times\mathcal{R}^{3}\,.
\end{align}
Similarly, the second-order tensors $\boldsymbol{A}$ and $\boldsymbol{B}$, or $A_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}$ and $B_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}$ respectively, have a scalar product given by
\begin{equation}
\boldsymbol{A}:\boldsymbol{B}=A_{ij}B_{ij} \,,
\end{equation}
an inner product given by
\begin{align}
\boldsymbol{A}\boldsymbol{B}&=A_{ij}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j})B_{k\ell}(\boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell}) \\
&=A_{ij}B_{j\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{\ell})\,,
\end{align}
and an outer product, also denoted by $\otimes$, that yields a fourth-order tensor $\mathbb{C}$ given by
\begin{align}
\mathbb{C}&=\boldsymbol{A}\otimes\boldsymbol{B}\\
&=A_{ij}B_{k\ell}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell}\\
& = \mathbb{C}_{ijk\ell} \, \boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell} \qquad \mathbb{C}\in\mathcal{R}^{3}\times\mathcal{R}^{3}\times\mathcal{R}^{3}\times\mathcal{R}^{3} \,.
\end{align}
Finally, the product of a fourth-order tensor $\mathbb{A}$ and a second-order tensor $\boldsymbol{B}$ is defined as
\begin{align}
\mathbb{A}\boldsymbol{B}&=\mathbb{A}_{ijk\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell})B_{mn}(\boldsymbol{e}_{m}\otimes\boldsymbol{e}_{n})\\
&=\mathbb{A}_{ijk\ell}B_{k\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j})\,,
\end{align}
The question is. If there is another tensor product, denoted by $\boxtimes$, and defined by
\begin{align}
(\boldsymbol{A}\boxtimes\boldsymbol{B})(\boldsymbol{a}\otimes\boldsymbol{b}) &= \boldsymbol{A}\boldsymbol{a}\otimes\boldsymbol{B}\boldsymbol{b} \,\text{, or} \\
(\boldsymbol{A}\boxtimes\boldsymbol{B})\boldsymbol{C} &= \boldsymbol{A}\boldsymbol{C}\boldsymbol{B}^{T}
\end{align}
how do the products $\boldsymbol{A}\otimes\boldsymbol{B}$ and $\boldsymbol{A}\boxtimes\boldsymbol{B}$ differ from each other? What do they represent physically? And, how would the product $\boldsymbol{A}\boxtimes\boldsymbol{B}$ be expressed in index notation?
 A: I think there is some confusion about the distinction between second order tensors and linear maps. On the one hand, your definition of $A\otimes B$ seems to imply that $A=A_{ij}e_i\otimes e_j$, which would imply that $A\in V\otimes V$, $V$ being the vector space in which the vectors live. On the other hand, in your definition of $A\boxtimes B$ you seem to treat $A$ as a linear mapping, in which $A$ would instead be an element of $V\otimes V^*\approx Hom(V,V)$. Of course if $V$ is finite dimensional, then there is an isomorphism between $V$ and $V^*$ anyway, but it can make things conceptually clearer to distinguish vectors from dual vectors.
With that said, let us consider $A$ and $B$ to be linear mappings $V\to V$, in which case $A\boxtimes B$ is an element of $Hom(V\otimes V, V\otimes V)$ and we have a sequence of isomorphisms:
\begin{eqnarray*}
Hom(V\otimes V, V\otimes V) &\approx & (V\otimes V)\otimes (V\otimes V)^*\\
&\approx & (V\otimes V)\otimes (V^*\otimes V^*)\\
&\approx & (V\otimes V^*)\otimes (V\otimes V^*)\\
&\approx Hom(V,V)\otimes Hom(V,V)\\
\end{eqnarray*}
and it is straightofrward to see that the image of $A\boxtimes B$ under this sequence is isomorphisms is precisely $A\otimes B\in Hom(V,V)\otimes Hom(V,V)$.
To see this more explicitly, let $e_i$ denote a basis of  $V$, with $e_i^*$ denoting the dual basis. Then the element of $V\otimes V^*$ corresponding to $A$ is given by $A_{ij} e_i\otimes e_j^*$. Note that here and in the rest of this answer, we will omit the explicit summation symbol over repeated indices.
On the other hand, we have
$$(A\boxtimes B) (e_k\otimes e_l)=Ae_k\otimes Be_l=A_{ik}e_i\otimes B_{jl}e_j=A_{ik}B_{jl} e_i\otimes e_j$$
Therefore we can trace through the image of $A\boxtimes B$ under the sequence of isomorphisms above as follows:
\begin{eqnarray*}
A\boxtimes B&\mapsto &A_{ik}B_{jl}(e_i\otimes e_j)\otimes(e_k\otimes e_l)^*\\
&\mapsto &A_{ik}B_{jl}(e_i\otimes e_j)\otimes(e_k^*\otimes e_l^*)\\
&\mapsto & A_{ik} (e_i\otimes e_k^*)\otimes B_{jl} (e_j\otimes e_l^*)\\
&\mapsto & A\otimes B
\end{eqnarray*}
