For functions of one real variable $t$ they call a critical point $t_0$ a saddle point if $t\mapsto f(t)-f(t_0)$ changes sign a $t_0$, and in most cases this means that the first nonvanishing derivative of $f$ at $t_0$ is of odd order.
For functions $f$ of $n\geq2$ real variables a critical point ${\bf p}=(p_1,\ldots, p_n)$ is a saddle point if the Hessian of $f$ at ${\bf p}$ is indefinite. This means that $f({\bf p}+{\bf X})-f({\bf p})={\bf X}'\ H\ {\bf X}+o(|{\bf X}|^2)\qquad({\bf X}\to{\bf 0})$ where the symmetric matrix $H=\left[{\partial^2 f(x_1,\ldots, x_n)\over \partial x_i\partial x_k}\right]_{i,k}$ has at least one strictly positive and at least one strictly negative eigenvalue. It follows that $f({\bf p}+{\bf X})-f({\bf p})$ takes as well positive as negative values in the immediate neighborhood of ${\bf p}$.
In the case of $n\geq2$ variables one usually assumes that the critical point in question is nondegenerate, meaning that the rank of $H$ is $=n$. The saddle points in the one-variable case are degenerate critical points in this sense, as the rank of the Hessian is $=0$ there.