These are all good questions
So, I'm guessing, the resultant focal length of the system (whatever that is), will also depend upon the distance between the lenses. How does the power of the lens fit into all this?
That is correct. In the limit where your lenses are thin the powers of two lenses is added as follows:
$$\phi_\text{tot} = \phi_1 + \phi_2 - \phi_1\phi_2\tau$$
where the individual powers are given by $\phi_1 = 1/f_1,\phi_2 = 1/f_2$, and $\tau = t/n$. $t$ is the spacing between the two lens elements and $n$ is the index of refraction of the medium between your two lenses ($n=1$ in vacuum). The effective focal length is then $f_\text{effective} = 1/\phi_\text{tot}$
There is a distinction between the effective focal length $f_\text{effective}$, the rear focal length $f_\text{R}$, and the front focal length $f_\text{F}$.
$$f_\text{R} = n_\text{R}f_\text{effective}\quad \quad f_\text{Front} = -n_\text{Front}f_\text{effective}$$
I.e. the front focal length is measured from the front principal plane to the front focal point (negative distance for a positive effective focal length) and the rear focal length is measured from the rear principal plane to the rear focal point (positive distance for a positive effective focal length).
Also, what do they mean by the 'resultant power of the lenses' in the first place? The resultant system of lenses is not going to be like these lenses which have negligible thickness. Where are you going to measure the focal length from?
This is a very important point. For a system of multiple lenses you measure the front and rear focal lengths from what are called the front and rear principal planes (rather than measuring directly from a lens element). For the case of a single lens the front and rear principal planes are located at the lens. For two lenses, the front principal plane is shifted from the first lens element a distance
$$d_\text{F} = + \frac{\phi_1}{\phi_\text{tot}} \frac{n_\text{F}\;t}{n}$$
the rear principal plane is shifted from the second lens element a distance.
$$d_\text{R} = -\frac{\phi_1}{\phi_\text{tot}}\frac{n_\text{R}\;t}{n}$$
For the above two equations I am defining the first lens to be on the left and the second lens on the right. The separation is $t$ and the index of the medium between the two lenses is $n$. The index of the medium to the right of the two lenses is $n_\text{R}$ and the index of the medium to the left of the two lenses is $n_\text{F}$. A value of $d_\text{F}$ or $d_\text{R}$ less than zero indicates a shift of the respective principal plane to the left; a value greater than zero indicates a shift to the right.
Disclaimer: this response is only a very brief introduction to the ideas of gaussian optics and the cardinal points. This stuff can be really confusing for anybody, especially when you consider all of the sign conventions. Also, these equations are really only valid in the paraxial limit for rotationally symmetric systems. Having said that, these basic formulations can be expanded to a system of any number of lenses -- not just 2. If you really want to understand this stuff there are surely some good books on this material. Hecht seems to be the book for intro optics, although I haven't read him. Check the table of contents to make sure gaussian optics is covered (it should be) because that's quite an expensive text.
The superposition is only approximately correct and the easiest way to understand both (1) why it works and (2) when it can be applied is to think in terms of waves and wavefronts, not rays. Since your link isn't working, let's write the equation down:
$$P_{lens}\approx \frac{n_{lens}-n_0}{n_o}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
With the fields thought of as waves, the lens surfaces become phase masks and the superposition of the two lens curvatures holds because the phase delays imparted by the phase masks one after the other are additive as long as the wavefront curvature (i.e. lateral phase distribution of the field) does not change much between the phase masks.
The power of the lens is the reciprocal of its focal length $f$, i.e. it is the reciprocal of the radius $f$ of curvature of the wavefronts that are output from the lens when a plane wave is input. A spherical wave of this radius converges to its diffraction limited focus after having propagated through this distance. So think of a plane wave input: the first surface represents a phase mask with phase delay as function of distance $r$ from the optical axis given by:
$$\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2 R_1}$$
the phase mask function for the second is:
$$-\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2 R_2}$$
and so the total phase delay is simply:
$$\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2} \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
Now ask yourself: what is the radius of curvature of the spherical wave that shows this (to terms of second order and lower) phase distribution. You'll find it is the reciprocal of the power given by the lensmaker's formula.
I give the full details of these kinds of calculations in this answer here. I show the calculation above as well as the transformer matrix method where one thinks of lens surfaces and other optical processing elements as operators in the group $SL(2,\,\mathbb{R})$.
Best Answer
We have to consider the individual refractive surfaces only and not the planoconvex lens as a whole, but it so happens, that the final answer can be simply expressed in terms of the focal length the planoconvex lens would have without the silvering.
For a general derivation, consider a lens with radius ($R_1,R_2$), with $R_2$ silvered. Cartesian sign convention are used here.
For an object distance (from the first surface, but for a thin lens, the distance is same for both the surfaces) $u$, we get for the first surface:- $$\frac{\mu}{v_1}=\frac{\mu-1}{R_1}+\frac{1}{u}$$ (where $\mu$ is the refractive index of lens and that of air is $1$)
Now for surface two, a mirror, $v_1$ is the object:- $$\frac1{v_2}=\frac2{R_2}-\frac{1}{v_1}$$ Now the second image $v_2$ acts as the object for first surface again, but now with inverted sign conventions. $$-\frac 1v=\frac{1-\mu}{-R_1}+\frac{\mu}{v_2}$$
using the value of $v_1$ and $v_2$ from preceding equations you come up with $$\frac 1v+\frac 1u=\frac{2\mu}{R_2}-\frac{2(\mu-1)}{R_1}$$ This is similar to a mirror with focal length given by the left hand side of the equation. It can be rearranged to $$\frac 1f=-\bigg(\frac{2(\mu-1)}{R_1}-\frac{2(\mu-1)}{R_2}\bigg)+\frac 2{R_2}=\frac{-2}{f_{\text{lens}}}+\frac{1}{f_{\text{mirror}}}$$ which is the required equation.