你需要加入 address 两次使用不同别名的表

select staff.address_id, a_staff.address as staff_adress,
       store.address_id, a_store.address as store_adress
from store
join staff on staff.store_id = store.store_id
join address a_staff on staff.address_id = a_staff.address_id
join address a_store on store.address_id = a_store.address_id

因为您用相同的地址联接staff和store表（实际上您将store表的address表和staff表与store联接，所以此结果集中的address是针对store而不是针对staff的）表，并且联接条件仅为store而不是针对staff的pick addresses和与相关store联接的staff表，所以为什么在结果集中得到相同的地址两列。