Form a free-body diagram and solve for the external reactions giving,

     ΣM_A = 0
     L B_y - F L/2 = 0
     B_y = F/2 = 300/2 = 150 lb

     ΣF_y = 0
     A_y + B_y - F = 0
     A_y = 300 - 150 = 150 lb

If the beam is cut just to the left of the applied load, then the proper free-body diagram for the section to the left of the cut is as shown. Notice, it does not include the 300 lb force since it is cut just before the force.

Sum the forces and moments to solve for the unknown internal loads:

     ΣF_y = 0
     A_y - V = 0
     V = A_y = 150 lb

     ΣM_A = 0
     M - V L/2 = 0
     M = LV/2 = 8 (150) / 2 = 600 lb-ft

The problem can also be solved by using the right section after making the cut. The free-body diagram for the right side section is as shown at the left. Notice, the 300 lb force is included on this section since the cut is to the left of the force. Summing the forces and moments give,

     ΣF_y = 0
     V + B_y - F = 0
    V = 300 - 150 = 150 lb

     ΣM_B = 0
     -V L/2 + F L/2 - M = 0
     M = -150 (8/2) + 300 (8/2) = 600 lb-ft

As expected, both methods obtain the same results, a shear force of 150 lb and a bending moment of 600 lb-ft.