Solution:

For ideal pulleys, the product of tension and acceleration remains constant:

$$T_1 a = T_2 a_2$$

Since $T_2 = 2T_1$, substitute to get:

$$T_1 a = 2T_1 a_2 \Rightarrow a_2 = \frac{a}{2}$$

$\boxed{\text{So, } M_2 \text{ accelerates upward with } \frac{a}{2}}$