Force due to surface tension in calculation of excess pressure inside a liquid drop

The surface acts like a stretched membrane. The attachment point of that membrane is the triple junction between the solid surface, the air, and the liquid surface. The summation of the forces involved at that triple point is what results in the contact angle of the liquid on the surface.

I think I understand your question. You wish to ask why the surface tension force acts only on the periphery (circumference) of the hemispherical water droplet. Well, it doesn't. It acts all over the surface, at every point. To understand the answer, draw any given line of a very small length (say $dx$) anywhere on the surface. The force $Tdl$ acts on it due to tension by its neighbouring molecules, say in a given direction (right, say). But it too exerts a net equal and opposite surface tension force $Tdl$ (towards the left). So the net force on the small line is zero. This argument can be extended to the whole surface. The only region where this line experiences a net non-zero force is the periphery. So the net force due to surface tension is exerted on the periphery as shown and is given by (assuming surface tension to be constant) $$F= int{Tdl} = T cdot 2pi R$$

Excess pressure inside a liquid drop

In the calculation you consider the balance of forces on a hemisphere of liquid. The only forces which are relevant are external forces which other objects exert on this hemisphere. These include the pull of surface tension $sigma$ from the other hemisphere at the rim where they join, and the pressure $p$ from the liquid inside the other hemisphere acting across the flat face of this hemisphere (the yellow area in the diagram). (Also the atmospheric pressure across the blue surface, which I have not shown. So in effect in my diagram $p$ is the excess pressure.)

It is not necessary to consider the surface tension forces at other points in the surface of this hemisphere (the blue area in the diagram) because these are internal forces. They always occur in equal and opposite pairs between adjacent elements of water which are both in this hemisphere. It is only along the circular rim of the hemisphere that the opposite paired surface tension forces are missing because they act on water which is in the other hemisphere not this one.