I tried to come up with proofs ($1st, 2$nd$ proofs are deleted) but got stuck in some point.
Seeing all problems, I come up with the third proof.
Let $f=T_2^{-1}T_1$, let $Q^n$ be the standard simplex in $\mathbb{R}^n$, $\sigma_0$ be the standard positively oriented n-simplex in $\mathbb{R}^n$, now $f(Q^n)=Q^n$.
Proof ($3$rd):
$f$ is a diffeomorphism on $Q^n$. Since $f$ is diffeomorphism, the topological boundary of $Q^n$ is mapped to itself.
Now it is defined that $\partial f=f(\partial \sigma_0)$ is the positively oriented boundary.
Since jacobian of $f$ $\gt 0$, $f(\sigma_0)$ have same orientation as $\sigma_0$, $f$ preserves orientation also on its boundary. Hence $\partial f= \partial \sigma_0$.
Edit: All previous proofs are deleted as I thought they are either erroneous or a rephrase of the $3$rd proof. Turn out, the question that remains is the consistency of all definition of orientation used, but that only means studies of differential geometry awaits.
Status: For proof verification.