This is an answer to the modified question, although, as @KentaSuzuki [suggested](https://mathoverflow.net/a/468376), 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 $k$, and the answer is the same.  I'll work in that generality.)  A parabolic subgroup of a reductive group does always have a Levi component, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is not a parabolic subgroup.

Of course, in this case, $P'$ *does* have a Levi component, but it's worth saying why.  Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, *is* a parabolic subgroup of $G$ [BT, Proposition 4.4], and its Levi component $M$ containing $T$ is also a Levi component of $P'$ [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.

[BT]:  [Borel and Tits - Groupes réductifs](http://www.numdam.org/item/?id=PMIHES_1965__27__55_0)

[1]:  This may be verified over the algebraic closure, so assume that $k$ is algebraically closed.  Let $U'$ be the unipotent radical of $P'$.  Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$.  Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.