I have seen the tensor product written as
$$
\left( \begin{array}{c}
a \\
b \\
\end{array} \right) \otimes
\left( \begin{array}{c}
c \\
d \\
\end{array} \right) = \left( \begin{array}{c}
ac \\
ad \\
bc \\
bd \\
\end{array} \right)$$
However I have also seen it written as
$$
\left( \begin{array}{c}
a \\
b \\
\end{array} \right) \otimes
\left( \begin{array}{c}
c \\
d \\
\end{array} \right) =
\left( \begin{array}{c}
a \\
b \\
\end{array} \right)
\left( \begin{array}{c}
c & d \\ \end{array} \right) =
\left( \begin{array}{c}
ac & ad\\
bc & bd \\\end{array} \right)
$$
Which I have seen in the context of outer products.
Why are there two ways to do a tensor product?
Best Answer
Both are different realizations of the tensor product.
Consider $V=\mathbb{R}^ 2$. Your first "equality" arises from the isomorphism
$$V \otimes V \to \mathbb{R}^4$$ $$e_1 \otimes e_1 \mapsto e_1;$$ $$e_1 \otimes e_2 \mapsto e_2;$$ $$e_2 \otimes e_1 \mapsto e_3;$$ $$e_2 \otimes e_2 \mapsto e_4.$$
The second "equality" arises from the isomorphism
$V \otimes V \simeq V \otimes V^* \simeq Hom(V,V)$, where the first isomorphism comes from the isomorphism of the dual with the space arising from the standard inner product of $\mathbb{R}^2$ (which is essentially just transposition), and the second one comes from $(w,v^*) \mapsto v^*(\cdot) w.$ Note that it becomes a $2 \times 2$ matrix in the end, which is exactly (not exactly, but represents canonically) an element of $Hom(\mathbb{R}^2, \mathbb{R}^2)$.
This is mathematically speaking. I don't know why one is more appropriate than the other in your context. What I can say is that the second way is very useful, because it allows us to translate an endomorphism in terms of something structurally and algebraically rich such as the tensor product. The first one seems to be simply a down to earth immediate way to realize the tensor product as an array.