\begin{align} f(x) &= a + b (x - x_0) + c (x - x_0)^2 + d (x - x_0)^3 \tag{1}\label{1} . \end{align}
Note that \eqref{1} defines $y$ as a function of $x$. Combined with two values of $x$, $x_0$ and $x_3$, we have all we need to make a conversion to equivalent 2D cubic Bezier segment, defined by its four control points
\begin{align} P_0(x_0,y_0),\,&P_1(x_1,y_1),\,P_2(x_2,y_2),\,P_3(x_3,y_3) \tag{2}\label{2} . \end{align} The end points are
\begin{align} P_0&=(x_0,y_0)=(x_0,f(x_0))=(x_0,a) \tag{3}\label{3} ,\\ P_3&=(x_3,y_3)=(x_3,f(x_3)) \tag{4}\label{4} . \end{align}
2D cubic Bezier segment is defined as usual,
\begin{align} B_3(t)&=(x(t),y(t)) \tag{5}\label{5} ,\\ x(t)&=x_0\,(1-t)^3+3\,x_1\,(1-t)^2\,t+3\,x_2\,(1-t)\,t^2+x_3\,t^3 \tag{6}\label{6} ,\\ y(t)&=y_0\,(1-t)^3+3\,y_1\,(1-t)^2\,t+3\,y_2\,(1-t)\,t^2+y_3\,t^3 ,\quad t\in[0,1] \tag{7}\label{7} . \end{align}
We know that $x$ is linear in $t$, so we must have \begin{align} x_0\,(1-t)+x_3\,t &=x_0\,(1-t)^3+3\,x_1\,(1-t)^2\,t+3\,x_2\,(1-t)\,t^2+x_3\,t^3 \tag{8}\label{8} ,\\ (x_3-x_0)t+x_0 &=(3x_1-x_0-3x_2+x_3)t^3+(3x_0+3x_2-6x_1)t^2+3(x_1-x_0)t+x_0 \quad \forall t\in[0,1] \tag{9}\label{9} , \end{align}
hence we have $x_1,x_2$ evenly distributed between the endpoints $x_0$ and $x_3$:
\begin{align} x_1 &= \tfrac13 (2x_0+x_3)=x(t)\Big|_{t=1/3} \tag{10}\label{10} ,\\ x_2 &= \tfrac13 (x_0+2x_3)=x(t)\Big|_{t=2/3} \tag{11}\label{11} . \end{align}
Corresponding $y$ points on the curve \eqref{1} are
\begin{align} y(\tfrac13)&=f(x_1) \tag{12}\label{12} ,\\ y(\tfrac23)&=f(x_2) \tag{13}\label{13} . \end{align}
The last pair of equations is a linear system with two unknowns, $y_1$ and $y_2$ which can be trivially solved as
\begin{align} y_1 &= \tfrac13\,b\,(x_3-x_0)+a \tag{14}\label{14} ,\\ y_2 &= \tfrac13(x_3-x_0)(c(x_3-x_0)+2b)+a \tag{15}\label{15} . \end{align}
Example
\begin{align} a &= 7 ,\quad b = 2 ,\quad c = 2 ,\quad d = -1 \tag{16}\label{16} ,\\ x_0 &= -1 ,\quad x_3 = 3 \tag{17}\label{17} ,\\ y_0&=a=7 ,\quad y_3=f(3)=-17 \tag{18}\label{18} ,\\ x_1 &= \tfrac13(2x_0+x_3)=\tfrac13 \tag{19}\label{19} ,\\ x_2 &= \tfrac13(x_0+2x_3)=\tfrac53 \tag{20}\label{20} ,\\ y_1 &= \tfrac13\,b\,(x_3-x_0)+a =\tfrac{29}3 \tag{21}\label{21} ,\\ y_2 &= \tfrac13(x_3-x_0)(c(x_3-x_0)+2b)+a =23 \tag{22}\label{22} . \end{align}
