Beaker and suspended ball weight measured by spring balance

When the ball is suspended, the weight of the ball is balanced by the tension in the string and the buoyancy force. Here buoyancy force on the ball has an equal and opposite reaction on the water itself thus this extra force must be accounted for.

Why shouldn't it be weight of beaker + Weight of ball - Bouyant force on the ball

This would be the case if the ball was not suspended and is at equilibrium at the bottom of the container . The apparent weight of the ball will be "Weight of ball - Bouyant force on the ball" The apparent weight + the container weight is the result you have arrived to.

Newton's Third Law: action begets reaction.

As the water exerts a buoyant force on the ball, the water exerts an opposite force on the beaker, thus scales. It's a classic part of density determination for irregularly shaped bodies.

No matter how much denser the ball is than the water, when held submerged but not sank to the bottom, the weight of the beaker on a scale will only increase by the weight of the volume of water the ball displaces, its buoyant force, per Archimedes principle of buoyancy. The tension on the string will support the extra amount of the ball's weight that is greater than that of the water displaced. You can see that the water level rises when the ball is submerged. It rises by the amount it would rise if the same volume of water as the ball's volume was added instead of the ball.