Quoting Taylor page 438 (the part that fixed it for me):
"A bead of mass m is threaded on a frictionless wire that lies in the xy plane (y vertically up), bent in the shape y = f (x) with a minimum at the origin, as shown in Figure 11.12. [just a general function in the XY that has some dip at the origin where the bead can oscillate back and forth] Write down the potential and kinetic energies and their simplified forms appropriate for small oscillations about 0.
This system has just one degree of freedom, and the natural choice of generalized coordinate is just x. With this choice, the potential energy is simply ##U=mgy=mgf(x)##. When we confine ourselves to small oscillations, we can Taylor expand ##f(x)##. Since ##f(0)=f^{ \prime }(0)=0##, this gives ##U(x)=mgf(x)\approx \frac { 1 }{ 2 } mg{ f(x) }^{ \prime \prime }x^{ 2 } ##
The kinetic energy is ##T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })##, where, by the chain rule, ##\dot { y } =f^{ \prime }(x)\dot { x }##. Therefore, ##T=\frac { 1 }{ 2 } [1+{ f }^{ \prime }(x)^{ 2 }]\dot { x } ^{ 2 }##. Notice that the exact expression for T depends on ##\dot{x}## as well as ##x##. However, since T already contains the factor ##\dot{x}^2##, when we make the small-oscillation approximation we can simply replace the term ##f^{\prime}(x)## by its value at x=0 (namely zero) and we get ## T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })\approx\frac { 1 }{ 2 } m\dot { x } ^{ 2 }## As expected, the small-oscillation approximation has reduced both U and T to homogeneous quadratic functions of ##x## (for U) or ##\dot{x}## (for T)."