In a Bézier curve, $x$ and $y$ are polynomials in the parameter $t$. Note that you can't just have "a part of the sine function": if $y(t) = \sin(x(t))$ for $t$ in some interval, since both sides of that equation are analytic functions on the complex plane the equation would be true for all complex numbers $t$. Since $y(t)$ is a polynomial, for any given value of $y$ (unless $y$ is constant) there are only finitely many $t$ and thus finitely many $x$. But this is not the case for the sine function: $\sin(n \pi) = 0$ for all integers $n$. So the sine curve can't be given exactly by a Bézier curve of any degree.
Solved it myself, as below. I don't feel that this solution is "right" somehow, because it treats the interior points $\mathbf B$ and $\mathbf C$ very asymmetrically, so improvements are invited. But, anyway ...
Let's use the notation from the last few paragraphs of the question.
It's clear that $\mathbf P(0) = \mathbf A$, and $\mathbf P(1) = \mathbf D$, so these two points are interpolated, already, and we only have to worry about the other two points, $\mathbf C$ and $\mathbf D$.
First, we find numbers $h$ and $k$ such that
$$\mathbf C = \mathbf A + h(\mathbf B - \mathbf A) + k(\mathbf D - \mathbf A)
= (1-h-k)\mathbf A + h\mathbf B + k\mathbf D$$
This is possible provided that $\mathbf A$, $\mathbf B$, $\mathbf D$ are not collinear. Then, since $\mathbf P(v) = \mathbf C$, we have
$$(1-h-k)\mathbf A + h\mathbf B + k\mathbf D =
(1-v)^2 \mathbf A + 2v(1-v) \mathbf P + v^2 \mathbf D $$
But, since $\mathbf P(u) = \mathbf B$, we know that
$$ \mathbf B = (1-u)^2 \mathbf A + 2u(1-u) \mathbf P + u^2 \mathbf D$$
Substituting for $\mathbf B$ on the left-hand side, and equating coefficients of $\mathbf A$, $\mathbf P$, $\mathbf D$ gives
$$(1-v)^2 = 1 - h - k +h(1-u)^2 $$
$$2v(1-v) = 2hu(1-u)$$
$$v^2 = hu^2 + k $$
We can easily eliminate $v$ from these last three equations using the fact that $[2v(1-v)]^2 = 4[v^2][(1-v)^2]$. We get:
$$[2hu(1-u)]^2 = 4[hu^2 + k][1 - h - k +h(1-u)^2]$$
After a little algebra, this reduces to:
$$h(1-h)u^2 - 2hku + k(1-k) = 0$$
So, we solve this quadratic for $u$, and then get the unknown interior control point $\mathbf P$ from
$$ \mathbf P = \frac{\mathbf B - (1-u)^2 \mathbf A -u^2 \mathbf D}{2u(1-u)}$$
The number of solutions depends on the number of real solutions of the quadratic. Quite often, you can draw two different quadratic Bezier curves through the four given points, as explained in the paper cited in the question.
Best Answer
The equation of the curve is $$ \mathbf{C}(t) = (1-t)^2 \mathbf{P}_0 + 2t(1-t)\mathbf{P}_1 + t^2\mathbf{P}_2 $$ If we let $\mathbf{P}_0 = (x_0,y_0)$, $\mathbf{P}_1 = (x_1,y_1)$, $\mathbf{P}_2 = (x_2,y_2)$, then we can write two separate equations for $x$ and $y$: $$ x(t) = (1-t)^2 x_0 + 2t(1-t)x_1 + t^2 x_2 $$ $$ y(t) = (1-t)^2 y_0 + 2t(1-t)y_1 + t^2 y_2 $$
You can rearrange and collect powers of $t$, if you want to: $$ x(t) = (x_0 - 2x_1 + x_2)t^2 + 2(x_1 - x_0)t + x_0 $$ $$ y(t) = (y_0 - 2y_1 + y_2)t^2 + 2(y_1 - y_0)t + y_0 $$ As you mentioned, it is not always possible to write the curve in the form $y=f(x)$. For example, the curve $$ x(t) = (t - \tfrac12)^2 \quad ; \quad y(t) = t - \tfrac12 $$ has equation $x = y^2$. We could write $y = \sqrt x$, but this is only half of the curve (the other half is $y = -\sqrt x$).
However, you can always write a quadratic Bezier curve in the form $g(x,y)=0$, where $g$ is some second-degree function of $x$ and $y$.