Why can we ignore the $x\frac{dU}{dt}$ term in the Taylor expansion if it is zero only at the equilibrium but not at any other point?

(Note in this expansion, it's $\frac{dU}{dx}$, not $\frac{dU}{dt}$-- we are expanding in $x$.) It's the definition of the Taylor expansion: if you pick a point $x_0$ to ``expand around'', the expansion is $U(x) = U_0 + x\left(\frac{dU}{dx}\right)_{0} + \frac{x^2}{2!}\left(\frac{d^2U}{dx^2}\right)_0 + ...$, where each derivative is evaluated at $x_0$. The Taylor expansion is valid for any chosen point $x_0$. If $x_0$ is chosen to be the equilibrium point, the first derivative vanishes.