This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).
One should be careful speaking casually of the Levi component of a subgroup, since linear algebraic groups over general fields don't have to have them. (You're working over $\mathbb C$, but the same question could be asked over any field, and the answer is the same.) A parabolic subgroup of a reductive group does always have a Levi component, but $P_1 \cap w P_2 w^{-1}$ is not a parabolic subgroup.
Of course, in this case, $P_1 \cap w P_2 w^{-1}$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $(P_1 \cap w P_2 w^{-1})U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$, and its Levi component $M$ containing $T$ is also a Levi component of $P_1 \cap w P_2 w^{-1}$. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected.