A combination of two reasons:
- Newton method attracts to saddle points;
- saddle points are common in machine learning, or in fact any multivariable optimization.
Look at the function $$f=x^2-y^2$$
If you apply multivariate Newton method, you get the following. $$\mathbf{x}_{n+1} = \mathbf{x}_n - [\mathbf{H}f(\mathbf{x}_n)]^{-1} \nabla f(\mathbf{x}_n)$$
Let's get the Hessian: $$\mathbf{H}= \begin{bmatrix} \dfrac{\partial^2 f}{\partial x_1^2} & \dfrac{\partial^2 f}{\partial x_1\,\partial x_2} & \cdots & \dfrac{\partial^2 f}{\partial x_1\,\partial x_n} \\[2.2ex] \dfrac{\partial^2 f}{\partial x_2\,\partial x_1} & \dfrac{\partial^2 f}{\partial x_2^2} & \cdots & \dfrac{\partial^2 f}{\partial x_2\,\partial x_n} \\[2.2ex] \vdots & \vdots & \ddots & \vdots \\[2.2ex] \dfrac{\partial^2 f}{\partial x_n\,\partial x_1} & \dfrac{\partial^2 f}{\partial x_n\,\partial x_2} & \cdots & \dfrac{\partial^2 f}{\partial x_n^2} \end{bmatrix}.$$
$$\mathbf{H}= \begin{bmatrix} 2 & 0 \\[2.2ex] 0 & -2 \end{bmatrix}$$
Invert it: $$[\mathbf{H} f]^{-1}= \begin{bmatrix} 1/2 & 0 \\[2.2ex] 0 & -1/2 \end{bmatrix}$$
Get the gradient: $$\nabla f=\begin{bmatrix} 2x \\[2.2ex] -2y \end{bmatrix}$$
Get the final equation: $$\mathbf{\begin{bmatrix} x \\[2.2ex] y \end{bmatrix}}_{n+1} = \begin{bmatrix} x \\[2.2ex] y \end{bmatrix}_n -\begin{bmatrix} 1/2 & 0 \\[2.2ex] 0 & -1/2 \end{bmatrix} \begin{bmatrix} 2x_n \\[2.2ex] -2y_n \end{bmatrix}= \mathbf{\begin{bmatrix} x \\[2.2ex] y \end{bmatrix}}_n - \begin{bmatrix} x \\[2.2ex] y \end{bmatrix}_n = \begin{bmatrix} 0 \\[2.2ex] 0 \end{bmatrix} $$
So, you see how the Newton method led you to the saddle point at $x=0,y=0$.
In contrast, the gardientgradient descent method will not lead to the saddle point. The gardientgradient is zero at the saddle point, but a tiny step out would pull the optimization away as you can see from the gradient above - its gradient on y-variable is negative.
