The number of local extremum points of $s(t)$ can be bound as a function of $n$ (where $n+1$ is the number of interpolation points).
The final bound, which I will develop using B-Splines, is $n-1$. I'll also show that his bound is tight.
First, note that there is a naive bound of $2n$ extremum points since each segment of the spline between the knots is just a cubic polynomial which has at most two extremum points.
However, as I said above, we can do better using the theory of B-Splines.
In the reference in the question each $[t_i, t_{i+1}]$ interval is of unit length, so we will
represent $s(t)$ using the uniform B-Spline basis functions where $\{t_i\}$ are equally spaced knots placed a unit length apart.
For simplicity (and without loss of generality) $t_i=i$, however the arguments below can probably be extended to non-uniform knot vectors as well.
The function representation is now (see for example Piegl, Les; Tiller, Wayne, The $NURBS$ books., Berlin: Springer. xiv, 646 p. (1997). ZBL0868.68106. or here)
$$
s(t) = \sum_{k=0}^{n+2} P_k N_{k,3}(t)
$$
Where $N_{k,3}(t)$ are the cubic B-Spline basis functions derived from the knot vector and $P_k$ are the control points (in this case control coefficients since they are scalars).
The ordered control points compose the control polygon, which has geometric properties that we will be using.
Note that the number of control points in the B-Spline representation is $n+3$.
(I'll be using the terms control points and control polygon a bit freely here. They can be formulated more formally using a parametric B-Spline curve $(x(t)=t, y(t)=s(t))$ of the graph of the function but I omit the details).
Solving the B-Spline interpolation problem is standard (see for example, Chapter 9 of The NURBS Book, or here). In order to compute the control points you evaluates the basis functions at the knots for $n+1$ linear constraints and the natural end conditions $s''(t_0)=s''(t_n)=0$ give you the additional two constraints.
Once we have the B-Spline representation, we have $n+3$ coefficients $P_k$ and we can use their geometric properties to give our bound.
The first property we use is the simple form of the B-Spline derivative.
The derivative of a cubic B-Spline function is itself a B-Spline function (of degree 2) and has a nice representation.
For our uniform knots case the derivative function is just
$$
s'(t) = \sum_{k=0}^{n+1} (P_{k+1}-P_k) N_{k,2}(t).
$$
The second property we use is the variation diminishing property of B-Splines, which we apply to the derivative function. This means that the number of intersection points of any line with the function graph is smaller or equal to the number of intersections of this line with the control polygon.
Applying it to our context means that the $x$-axis does not intersect the graph of the derivative function more than $n+1$ times (since there are $n+1$ segments in a control polygon with $n+2$ control points).
Therefore, we have a bound of $n+1$ on the zero-crossing of the derivative.
Thus, there cannot be more than $n+1$ extremum points to the original function $s(t)$.
For the natural spline, we can in fact do slightly better.
The natural end-conditions imply that $P_1$ lies on the line between $P_0$ and $P_2$ (and similarly for the last end points).
From this follows that $P_1-P_0$ and $P_2-P_1$, the first control points of the derivative, have the same sign (and similarly for the last points).
Therefore, the first (and last) two segments of the derivative control polygon can only contribute at most one zero-crossing and therefore the maximal number of zero crossings is $n-1$.
So (at last), our final bound is $n-1$ extremum points.
This bound is tight; for any $n$ we can construct the following $n+1$ configuration of points for which the natural spline will have $n-1$ extremum points.
An alternating $(n+1)$-sized point sequence of $\pm 1$ will have $n-1$ extremum points (see the figure for an example with $n=10$). This can be proved by explicitly solving the interpolation equations but a simpler proof comes from the mean value theorem. By the mean value theorem every alternating segment $(t_i,-1), (t_{i+1}, 1)$ has a positive derivative of value $s'(c_i)=2=\frac{1-(-1)}{t_{i+1}-t_i}$, and its consecutive segment will have a negative derivative of value $s'(c_{i+1})=-2=\frac{(-1)-1}{t_{i+2}-t_{i+1}}$. Thus, by the continuity of the derivative we have a zero derivative in the interval $(c_0,c_1)$ and another one in $(c_1,c_2)$ and so on.. until the last interval $(c_{n-2},c_{n-1})$, for a total of $n-1$ zero derivatives.
Conclusion: From the theory of B-Splines and their geometric properties, it follows that
the maximal number of extremum points of $s(t)$ is $n-1$ and this bound is tight since for any $n$ we can construct a configuration of interpolation points for which this bound is attained.
