A $C^2$ piecewise Hermite interpolant and a cubic spline are one and the same!
Remember what's done to derive the tridiagonal system: we require that at a joining point, the second left derivative and the second right derivative should be equal.
To that end, consider the usual form of a cubic Hermite interpolant over the interval $(x_i,x_{i+1})$:
$$y_i+y_i^{\prime}\left(x-x_i\right)+c_i\left(x-x_i\right)^2+d_i\left(x-x_i\right)^3$$
where
$$\begin{align*}c_i&=\frac{3s_i-2y_i^\prime-y_{i+1}^\prime}{x_{i+1}-x_i}\\
d_i&=\frac{y_i^\prime+y_{i+1}^\prime-2s_i}{\left(x_{i+1}-x_i\right)^2}\\
s_i&=\frac{y_{i+1}-y_i}{x_{i+1}-x_i}\end{align*}$$
and $\{y_i^\prime,y_{i+1}^\prime\}$ are the slopes (derivative values) of your interpolant at the corresponding points $(x_i,y_i)$, $(x_{i+1},y_{i+1})$.
Take the second derivative of the interpolant over $(x_{i-1},x_i)$ evaluated at $x=x_i$ and the second derivative of the interpolant over $(x_i,x_{i+1})$ evaluated at $x=x_i$ and equate them to yield (letting $h_i=x_{i+1}-x_i$):
$$c_{i-1}-c_i+3d_{i-1}h_{i-1}=0$$
Replacing $c$ and $d$ with their explicit expressions and rearranging yields:
$$h_i y_{i-1}^{\prime}+2(h_{i-1}+h_i)y_i^{\prime}+h_{i-1} y_{i+1}^{\prime}=3(h_i s_{i-1}+h_{i-1} s_i)$$
which can be shown to be equivalent to one of the equations of your tridiagonal system when $h$ and $s$ are replaced with expressions in terms of $x$ and $y$.
Of course, one could instead consider the cubic interpolant in the following form:
$$y_i+\beta_i\left(x-x_i\right)+\frac{y_i^{\prime\prime}}{2}\left(x-x_i\right)^2+\delta_i\left(x-x_i\right)^3$$
where
$$\begin{align*}\beta_i&=s_i-\frac{h_i(2y_i^{\prime\prime}+y_{i+1}^{\prime\prime})}{6}\\\delta_i&=\frac{y_{i+1}^{\prime\prime}-y_i^{\prime\prime}}{6h_i}\end{align*}$$
Doing a similar operation as was done for the Hermite interpolant to this form (except here, one equates first derivatives instead of second derivatives) yields
$$h_{i-1} y_{i-1}^{\prime\prime}+2(h_{i-1}+h_i)y_i^{\prime\prime}+h_i y_{i+1}^{\prime\prime}=6(s_i-s_{i-1})$$
which may be the form you're accustomed to.
To complete this answer, let's consider the boundary condition of the "natural" spline, $y_1^{\prime\prime}=0$ (and similarly for the other end): for the formulation where you solve the tridiagonal for the second derivatives, the replacement is straightforward.
For the Hermite case, one needs a bit of work to impose this condition for the second derivative. Taking the second derivative of the interpolant at $(x_1,x_2)$ evaluated at $x_1$ and equating that to 0 yields the condition $c_1=0$; this expands to
$$\frac{3s_1-2y_1^\prime-y_2^\prime}{x_2-x_1}=0$$
which simplifies to
$$2y_1^\prime+y_2^\prime=3s_1$$
which is the first equation in the tridiagonal system you gave. (The derivation for the other end is similar.)
(1) $(x_1, \ldots, x_m)$, to get the right dimension for the spline space (to match the number of constraints).
(2) It means that the cardinal b-splines have $L_\infty$ norm (maximum absolute value) that is independent of the grid size, $h$. I'm not sure why this would be the case. Need more context to explain further.
(3) A cardinal spline is one that has the value 1 at one interpolation point, and zero at all others. So, it's a statement about the values of the spline function, not about its knots. So, it doesn't make sense to ask whether or not the splines you mentioned are cardinal splines, since you didn't say anything about values.
(4) Some time ago there was a "spline toolkit" in Matlab. The code was written by Carl deBoor, based on his book "A Practical Guide to Splines" (which I highly recommend). Looking at the Matlab web site, the spline toolkit seems to have disappeared, and they are now talking about a "Curve Fitting" package, instead. But, anyway, it looks like a lot of the original code/functions are still present. For tensor product interpolation, look at this document, or look at deBoor's book.
Best Answer
Consider the general $3$rd degree "basis spline" (sometimes called "bump spline") defined on the interval $[x_0-2h,x_0+2h]$ in four parts in the following way:
$$\varphi(x)=\begin{cases}\tfrac16(2h+(x-x_0))^3& x \in [x_0-2h,x_0-h]\\ k-\tfrac12(x-x_0)^2(2h+(x-x_0))& x \in [x_0-h,x_0]\\ k-\tfrac12(x-x_0)^2(2h-(x-x_0))& x \in [x_0,x_0+h]\\ \tfrac16(2h-(x-x_0))^3& x \in [x_0+h,x_0+2h] \end{cases} \ where \ k:=2h^3/3.$$
In our case, $h=\tfrac12$ and $x_0=0$, giving simplified expressions:
$$\varphi(x)=\begin{cases}\tfrac16(x+1)^3& x \in [-1,-\frac12]\\ \tfrac{1}{12}-\tfrac12 x^2(x+1)& x \in [-\frac12,0]\\ \tfrac{1}{12}-\tfrac12 x^2(1-x)& x \in [0,\frac12]\\ \tfrac16(1-x)^3& x \in [\frac12,1] \end{cases}$$
with this graphical representation:
As $\varphi(0)=\tfrac{1}{12}$ and $\varphi(-1)=\varphi(1)=0$, take $$f(x)=12 \varphi(x) + 1$$ in order to comply with conditions $f(0)=2$ and $f(-1)=f(1)=1$.
Afterwards, replicate it in a periodic way by taking $x_0=2n$ for any integer $n$:
Remarks:
You can check that the continuity conditions of $f$, $f'$ and $f''$ are fulfilled at the "connections".
For explanations, and more generaly a nice presentation of splines, see this slideshow.