$\def\ZZ{\mathbb{Z}}$Your conjecture is right when you require $A$ to be primitive. The version where you set $(X,kY) \sim (kX,Y)$ doesn't work even in your chosen example.

$A$ gives a map $\ZZ^2 \to \ZZ^2$. Set $K = \ZZ^2/A \ZZ^2$; this is an abelian group of order $n$ generated by $2$ elements and the condition that $A$ is primitive implies that $G \cong \ZZ/n \ZZ$.

You want to know how many ways you can factor this as $\ZZ^2 \to L \to \ZZ^2$, where $L \cong \ZZ^2$ and we work up to isomorphisms on the middle factor. Such a factorization is uniquely determined by the subgroup $L/ZZ^2$ of $K$. So we are counting subgroups of $\ZZ/n \ZZ$, which there are $\sigma_0(n)$ of.

If we take $A = \left( \begin{smallmatrix} 2 & 0 \\ 0 & 2 \end{smallmatrix} \right)$, then $G \cong (\ZZ/2 \ZZ)^2$, with five subgroups. Representative factorizations are
$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \quad \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$$
$$\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad
\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1& 1 \\ 1 & -1 \end{pmatrix}.$$
Your equivalence relation collapses these $5$ cases to $4$, not $3$.

It is clear that $\#(T_A/G)$ will only depend on the isomorphism type of the abelian group $\ZZ^2/A \ZZ^2$ (in other words, on the Smith normal form of $A$). If you really need it, I could work it out; I suspect you could too.