Determine in which class $L=\big\{\langle M_1,M_2,w\rangle\mid M_1,M_2\text{ are TM and }L(M_1)\cap L(M_2)=\{w\}\big\}$
No. It is not correct.

For example, suppose that $x=0$, $w=00$, $L(M_1)\bigcap L(M_2)=\{x,w\}$.  If $M_1(x)$ or $M_2(x)$ takes more than $|x|=1$ steps to answer "yes", then your proposed algorithm will failed.