Breaking up the problem into two parts we can solve one part at a time. First we will examine the Forces on the TOP BLOCK.
It does not move in the y-plane so its weight is balanced by a normal force exerted by the lower block (The normal force on the block is equal to its weight, 30N). The only Net force that the top block feels is the static friction that the bottom block exerts on it (2). Solve for the mass of the top block in (3). Using the static frictional force and the mass of the top block, the acceleration can be found (4).
Clearly the top and bottom blocks must accelerate at the same rate (a of the top block equals a of the bottom block) so that they remain in contact. We move our attention to the lower block now. Again there is no movement in the y-plane, so the only net forces will come in the x-plane, but you have to remember the weight of the top block pushing on the lower block.
The normal force that the floor exerts on the bottom block is the weight of both blocks added together (6). This is important when you calculate the kinetic friction between the floor and the lower block (7). However when doing the 
calculations, you only use the mass of the lower block by itself (8)-(13) because you are treating the system as two separate blocks, even if they are moving at the same acceleration.
