Reparametrize t as u(s), as we did before. The derivative of p(u(s)) brings in u′ by the chain rule. Pull u′ out of the norm, giving f(p(u(s)))×|p′(u(s))|×u′(s). Now the entire integrand is merely a form of integration by substitution. We replaced t with u(s); that doesn't change the integral. Thus the answer depends only on the path and the field.
For a simple example, let the path be the unit circle, cos(t),sin(t). Imagine this is a loop of wire whose density is y2. The top and bottom are heavy while the left and right weigh almost nothing. How much does the wire weigh? Apply the path integral. The speed is 1, so we can ignore that. This leaves an integrand of sin2(t). The integral is ½(t-sin(2t)), from 0 to 2π, hence π. This is also the total heat contained in the wire if f gives the absolute temperature.
If we reverse the path, replacing t with -t, the line integral changes sign, but not the path integral. Although p′ is negated, the path integral takes its norm, and that destroys the sign. The change in energy is reversed when going up or down a hill, but the wire has the same mass whether it is measured clockwise or counterclockwise.