First: it is false that if $g\in G$ then $T(G)g = T(G)$, but I suspect you meant "$g\in T(G)$", which would make that statement true.
Second: it is false that if $g\notin T(G)$ then $T(G)g=G-T(G)$. It's true that the coset $T(G)g$ is disjoint from $T(G)$, and hence is contained in $G-T(G)$, but it need not be equal to $G-T(G)$. In general it won't be (it will only be the case if $T(G)$ is of index $2$).
Here's what you need to do: take an element of $G/T(G)$, which looks like $T(G)g$ with $g\in G$, and assume that it is a torsion element: that is, there exists $n\gt 0$ such that $(T(G)g)^n = T(G)$. This will occur if and only if $T(G)g^n = (T(G)g)^n = T(G)$, if and only if $g^n\in T(G)$.
Can you now prove that in fact $g\in T(G)$? If so, then $T(G)g = T(G)$, and you will have proven that:
If $T(G)g\in T(G/T(G))$, then $T(G)g = T(G)$.
which will show that $T(G/T(G)) = \{T(G)\}$ (which happens to be the identity of $G/T(G)$).