Heuristically, when $\sigma$ and $\rho$ are "close," we can write
$$\sigma=(1-\epsilon)\rho+\epsilon\varrho$$ for some small positive number $\epsilon$ and some other normalized state $\varrho$. Then
$$\sigma-\rho=\epsilon(\varrho-\rho),$$ so $T$ is the projector onto the positive eigenspace of $\varrho-\rho$.
If we take the extreme case that $\rho$ and $\varrho$ have completely different eigenspaces, then the positive eigenspace is just the eigenspace of $\varrho$. Then we can compute the two projected operators
$$T\sigma T=(1-\epsilon)T\rho T+\epsilon T\varrho T\approx0+\epsilon \varrho $$ and
$$T\rho T\approx 0,$$ so the projector is seen to be approximately orthogonal to both $\sigma$ and $\rho$. I don't know in what sense this can be written as the "opposite" of $\rho$.
If we take the more realistic case where the eigenspaces of $\rho$ and $\varrho$ can overlap slightly, through something like $\rho=\delta\varrho+(1-\delta)\varrho^\prime$ for small $\delta>0$, then the positive eigenspace of $\varrho-\rho$ is the positive eigenspace of $\varrho-\varrho^\prime$.
With orthogonal $\varrho$ and $\varrho^\prime$, $T$ again projects onto the eigenspace of $\varrho$, such that we now have $T\sigma T=(1-\epsilon)\delta\varrho+\epsilon \varrho$ and $T\rho T=\delta \varrho$. Again, the projector seems approximately orthogonal to both $\sigma$ and $\rho$ without being "opposite" to $\rho$ in any sense. In fact, it is impossible to have $T\rho T\approx - \rho$ because $T$ is a positive operator, so the word "opposite" is probably not defined in a common sense of the word.
