Converting comments to answer, as requested:
The Dandelin spheres answer question (1): a focus of a conic section is the point of tangency of its plane with one of those spheres. Clearly, the point on tangency lies on the cone axis if and only if the plane is perpendicular to that axis; therefore, the axis contains a focus in, and only in, the case of a circle. The axis point has some relation to the conic section, but not one as interesting or useful as a focus. (As for "Does the diameter of the base of the cone pass through the focus of the hyperbola?": note that the cone has no base. The thing extends, and expands, infinitely-far.)
This notion of points seeming to move "out to infinity and come back on the other side" is not uncommon in analytical geometry. (It's kinda what happens along asymptotes, when you think about it.) You might be interested in studying Projective Geometry, which adds a "line at infinity" to the standard Euclidean plane; this broader context helps unify all the conic sections into a single kind of curve that has various different relations to that line. (The parabola, in particular, has its "second vertex" on that line.)
One reason why your idea might not have panned out is that the point $B'$ in your question is the parabola’s vertex, so it doesn’t even work as the control point for a Bézier parameterization of the original parabola: even for a rational Bézier curve the middle control point never lies on the curve. The correct middle control point for a Bézier parameterization of the parabola can be found by intersecting the tangent planes to the cone at $A$ and $C$ with the plane that generates the parabola. We can thus find that with $$A = \left(2\sqrt{d(r-d)},r-2d,h\right) \\ B = \left(0,r,\frac hr(r-2d)\right) \\ C = \left(-2\sqrt{d(r-d)},r-2d,h\right)$$ the Bézier curve $$(1-t)^2A+2t(1-t)B+t^2C \tag1$$ matches the parabola. For $0\le t\le1$, we get the arc through the vertex with endpoints $A$ and $C$. (This orientation might not match your diagram exactly.)
Going back to your idea of projecting the parabola onto a variable plane through $A$ and $C$, we observe that since the center of the projection is the origin, if a point $P$ has a finite projection onto the plane then it is a scalar multiple of $P$ and so can be expressed as $P'=c(P)P$. Note that this scalar $c$ depends on $P$, and, of course, on the image plane. By construction, $A$ and $C$ are their own images, so the projection of the curve (1) is $$(1-t)^2A+2t(1-t)c(B)B+t^2C. \tag2$$ Comparing the rational Bézier parameterization $${(1-t)^2w_1P_1+2t(1-t)w_2P_2+t^2w_3P_3 \over (1-t)^2w_1+2t(1-t)w_2+t^2w_3}$$ to (2) suggests setting $w_1=w_3=1$ and since we’d like the control point to lie on the curve’s plane, $w_2=1/c(B)$ seems like a reasonable guess, leading to $${(1-t)^2A + 2t(1-t)B'/c(B)+t^2C \over (1-t)^2+2t(1-t)/c(B)+t^2}. \tag3$$
Parameterizing the variable plane by the angle $\phi$ between its normal and the positive $z$-axis, we have for an equation of the plane $$y\sin\phi+z\cos\phi=h\cos\phi+(r-2d)\sin\phi$$ and we can find $$B' = {r(h\cos\phi+(r-2d)\sin\phi) \over h(r-2d)\cos\phi+r^2\sin\phi}B.$$ I’ll leave verifying that using these values of $B'$ and $w$ do indeed produce a parameterization of the intersection curve, which involves some tedious but straightforward algebra.
Observe that, instead of sliding the vertex along the cone as you proposed, we are effectively sliding the parabola’s control point $B$ along the line through it and the origin. Parameterizing this family of curves by $c(B)$ as in (3) instead of the plane’s normal angle makes this relationship more transparent, I think.
Here, for example, is what results with $h=3$, $r=1$, $d=1/4$ and $\phi=\pi/4$:
The black arc is the above rational Bézier curve for $0\le t\le1$.
Best Answer
Consider this diagram of viewing the $x$-$y$ plane from this perspective:
The eye/retina of the viewer is a length of $d$ away from the plane perpendicular to the plane of the ground and passing through the $x$-axis. This plane is the plane in which the $x$-$y$ plane is being viewed. The point is also a height $h$ above the ground. The viewer/viewing plane has a coordinate system $x'$-$y'$.
The point $O' = (0,0)$ for the $x'$-$y'$ coordinates corresponds to the point $(0, \infty)$ in the $x$-$y$ coordinates. All other points for the viewer are below the viewer's $x'$ axis.
Using similar triangles we get that $\frac{x'}{d} = \frac{x}{y+d}$ and $\frac{y'}{d} = -\frac{h}{y+d}$ (the $-$ sign is because everything is below the $x'$ axis for the viewer).
Rearranging these equations we get $y'(y+d)=dh$ so $y = \frac{-d(y'+h)}{y'}$ and $x'y+dx'=dx$ so $x = \frac{x'y}{d}+x'$. Plugging the expression for $y$ into the expression for $x$ we get $x = x'(1-\frac{(y'+h)}{y'}) = \frac{-hx'}{y'}$.
Thus, the final transformations are $x = \frac{-hx'}{y'}$ and $y = \frac{-d(y'+h)}{y'}$.
The equation of a general conic is $Ax^2+By^2+Cxy+Dx+Ey+F = 0$.
Plugging these transformations in we get:
$h^2(A+dC)x'^2+d(Bd-E+1)y'^2+h(dC-D)x'y'+Ch^2dx'+hd(2Bd-E)y'+Bh^2d^2 = 0$
This is an equation of the form $A'x'^2+B'y'^2+C'x'y'+D'x'+E'y'+F' = 0$, which is also the equation of a conic. Thus, conics are mapped to conics under this transformation.
For example, the parabola $y = x^2$ is mapped to the ellipse $\frac{x'^2}{1/4}+\frac{(y'+\frac{1}{4})^2}{1/16} = 1$ for $d = 1$ and $h = \frac{1}{2}$ as shown in the image below.
The parabola $y=x^2$ under the transformation for $d = 1$ and $h = \frac{1}{2}$: