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.)
Best Answer
Let $p_3(x) = ax^3 + bx^2 + cx + d$ and without restricting generality suppose that $x_0 < x_1 < x_2$. Then in order for $p_3(x)$ to satisfy the given equations, its coefficients must satisfy the linear system $$ \begin{bmatrix} 1 & x_0 & x_0^2 & x_0^3\\ 1 & x_2 & x_2^2 & x_2^3\\ 0 & 1 & 2x_1 & 3x_1^2\\ 0 & 0 & 2 & 6x_1 \end{bmatrix} \begin{bmatrix} d\\ c\\ b\\ a \end{bmatrix} = \begin{bmatrix} f(x_0)\\ f(x_2)\\ f'(x_1)\\ f''(x_1) \end{bmatrix}, $$ where each row corresponds to one of the equations. Row reducing the coefficient matrix to a partial RREF we obtain $$ \begin{bmatrix} 1 & x_0 & x_0^2 & x_0^3\\ 0 & 1 & x_2+x_0 & x_2^2 + x_0x_2 + x_0^2\\ 0 & 0 & 2x_1-x_2-x_0 & 3x_1^2 - (x_2^2 + x_0x_2 + x_0^2)\\ 0 & 0 & 2 & 6x_1 \end{bmatrix} $$ From here it is easy to compute the determinant, which is $$ 6x_1^2 + 2x_2^2 + 2x_0^2 + 2x_0x_2 -6x_1x_2-6x_0x_1 = 3(x_1-x_0)^2 + 3(x_2 - x_1)^2 - (x_2-x_0)^2 $$ Now if you let $d_1 = x_1 - x_0 > 0$ and $d_2 = x_2 - x_1 > 0$, then $x_2 - x_0 = d_1 + d_2$. Thus the determinant may be written as $$ 3d_1^2 + 3d_2^2 - (d_1 + d_2)^2 = 2d_1^2 - 2d_1d_2 + 2d_2^2 = (d_1 - d_2)^2 + d_1^2 + d_2^2 > 0 $$ Since the determinant is nonzero, the linear system must have a unique solution. Hence, there exists a unique polynomial $p_3(x)$ that satisfies the given equations.