This derivation follows the line of the one at the bottom of page 109 in the text, except that the text's assumes that the starting velocity of the object (and so its starting kinetic energy) is zero. You'll see why the author did that when you compare the derivations...
Suppose a constant net force Fnet acts on an object of mass m over some distance . Newton's Second Law tells us that the object will have an acceleration a = Fnet/m, or:
Fnet = ma
It is a perfectly legal mathematical operation to multiply both sides of this equation by the distance the object moved, :
From kinematics, we know that:
so substituting this into the previous result gives:
Since and , we can get:
and combining like terms gives the Work-Energy Equation: