Why do we do right shifting in Booths algorithm?
Booth's recoding scheme can be thought of as follows. Instead of using the radix 4 canonical digit set \$\{0,1,2,3\}\$ for the multiplier, we recode the digits into new set \$\{-2,-1,0,1,2\}\$, which by the way is a redundant digit set but non-redundant can also be used. This is done simply because we do not like to multiply by 3 as it cannot be easily implemented in hardware. So, in your example $$Q = 0\ 11\ 10 = 1\ -1\ \ 2.$$ Then you can perform the multiplication as usual as you do in a radix 4 case.
Since we are only multiplying by \$\pm2\$ we only require a right shift for multiplication, and maybe a complementation if the multiplier digit is negative.