In method 1 block A moves in a straight line. In method 2 block A is fixed to and rotates with the disk. In both cases you have applied $\tau=I\alpha$.
Method 2 does not generally give you the correct answer. Your diagram suggests that block A is an extended object which is rigidly attached to the disk at one corner. However, as you have applied method 2, block A is a point particle concentrated at the point of attachment. Your method works only because you have treated block A as a point particle attached at the rim of the disk where the string makes contact.
If the disk and block A have moments of inertia $I_D$ and $I_A$ about their own centres of mass, and block A is attached rigidly to the disk with its centre of mass at a distance $b$ from the axis of rotation, then using the Parallel Axes Theorem the total moment of inertia of the composite object is $I=I_D+I_A+mb^2$
where $m$ is the mass of block A. Your method assumes that $b=r$, the radius of the disk, and $I_A=0.$ That is, you are assuming that block A is a point particle.
The method works for a point particle because, when such a particle is rotating in a circle it has the same kinetic energy as when translating in a straight line with the same speed. Your method does not work for an extended object which is rigidly attached to the disk because as it rotates around the centre of the disk an extended object also rotates about its own centre of mass. Unlike the block in method 1, it changes its orientation in space. In addition to its orbital KE is also has spin KE, so it has more KE than when translating at the same speed. Point particles have no spin KE even if they are spinning, because they have zero moment of inertia.
Method 2 would also give the same answer as method 1 if block A is attached to the rim of the disk at its centre of mass in such a way that block A is able to keep the same orientation in space as the disk turns. Then, like a point particle, it would have no spin KE, only orbital KE.