First of all, don't think of multipole moments as separate things that have their own individual meaning. Instead, think of them as parts of one thing. Once we have all the parts written down, we can start naming and organizing each one to determine its contribution to the whole.
Now, for your question
Is there a physical interpretation for multipole moments?
Yes! And they don't necessarily have to do anything with electrostatics, spherical harmonics or geometric series. A multipole expansion of some object in some basis is saying Hmmm, I have this funny shaped thing that isn't an elementary mathematical function, but I want to express it as a sum of elementary functions. (Fourier or someone said you can always do this with enough ink and parchment.)
So you first pick your basis, whether it be sine waves or exponentials or polynomials or the like, and then you start adding more and more terms of that basis, beginning with the lowest order (simplest) component.
For a non-mathematical example, consider this drawing of a cartoon sheep:
Step 1 is very simple since it's basically an oval with ripples. Right off the bat, with the very first thing we drew we have expressed some 75% of what the sheep will look like. This is important in multipole expansions: the lowest order terms dominate. If Step 1 were a square or a triangle then the whole sheep would likely be unrecognizable.
Step 2 does two things: it adds to the drawing and it slightly modifies some of what Step 1 did. You may have heard this called a corrective term or a higher order term. This would be the second term in your multipole expansion.
Step 3 adds less to the picture than Step 2, but look how far we've come. With just the first three multipoles, I'm betting a large percentage of people would recognize our animal as a sheep already. If instead of drawing a sheep with blobs we were constructing an E&M field with Legendre Polynomials, this is about where we stop since we have a good physical view of what's happening (this is the trade off of simplicity vs. accuracy present in all multipole expansions).
Additional steps just add more detail, filling out the gaps in the data at the cost of doing more work and keeping track of more pencil marks.
What is the physical interpretation of higher order multipole moments?
In E&M, we break the arbitrary looking charge distribution into multipoles with the hopes of drawing as much of the sheep as we need with as few details as possible. The multipoles look like:
0. the monopole, the offset that affects the E field in all directions the same way
1. the dipole, the description of how different two halves of the field would be if you drew a symmetry line right down the center
2. the quadrupole, a similar concept to the dipole, but instead of affecting two directions differently, you affect four directions
And you can keep going to as high of a multipole as you want (technically, you need to go to infinity to perfectly redraw an arbitrary sheep, but this isn't useful when we're just trying to predict system certain properties to within a finite precision to begin with).
Summary
A multipole expansion of anything is just breaking it down into a preferred basis. If we picked a good basis, we only need the first few multipoles because after that we're just touching up details we'll never need. Some multipoles are so useful we give them names, like the E&M charge distribution that affects everything isometrically (total charge) and antisymmetrically (dipole).
First understand the method of image charges. The idea behind the method is to bypass actually solving the differential equation with boundary conditions, and instead "cheat" by guessing the correct solution. To this end, we find a configuration of imaginary charges that together with the real ones will make the potential on all surfaces be what is given.
In your case, you have two surface, each with constant potential zero. For a plane, if you have a charge $q$ at $(x,y,z)$ and another one at $(x,-y,z)$, clearly the potential at $y=0$ will be zero. For the sphere, if a charge is at radius $R$ from the sphere with radius $a$, an image charge with charge $-qa/R$ should be placed at radius $a^2/R$, but the same idea remains.
Now, first add the image charge for the sphere. Now the potential on the sphere is zero, but on the plane, we still have a gradient. However, if you mirror both charges off the plane, you should see that the potential on both the sphere and the plane is zero.
Writing this all down, we have:
$$\begin{eqnarray}
\phi({\bf x}) &=& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y-R\sin\alpha)^2+z^2}} \\
&-& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y+R\sin\alpha)^2+z^2}} \\
&-&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y-(a^2/R)\sin\alpha)^2+z^2}} \\
&+&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y+(a^2/R)\sin\alpha)^2+z^2}}
\end{eqnarray}$$
Clearly, at $y=0$ we have $\phi=0$. Now, what about on the hemisphere? points on the hemisphere satisfy $z^2 = a^2 - x^2-y^2$, so that substituting leads us to the potential on the hemisphere:
$$\begin{eqnarray}
\phi({\bf x}_{sphere}) &=& \frac{q}{\sqrt{-2xR\cos\alpha -2yR\sin\alpha + R^2 +a^2}} \\
&-& \frac{q}{\sqrt{-2xR\cos\alpha +2yR\sin\alpha + R^2 +a^2}} \\
&-&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha -2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&+&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha +2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&=&0
\end{eqnarray}$$
By putting the $a/R$ into the square root.
To write down the multipole expansion, you just need to write down the Taylor expansion of the potential around $1/r$, with $r = \sqrt{x^2+y^2+z^2}$. This gives that any expression of the form:
$$\lim_{r\to\infty}\frac{1}{\sqrt{r^2+b}}= \frac{1}{r} - \frac{b}{2r^3} + \frac{3b^2}{8r^5} + \cdots$$
You can then use this expansion on the components of the potential to get your result.
Best Answer
The monopole moment is the total quantity of whatever it is you're considering. For example, if you're looking at a 3-dimensional multipole expansion of the mass of the earth, the monopole moment would be the total mass. That doesn't have anything to do with treating the earth like a sphere. Instead, it's the statement that the total mass of the earth (regardless of its shape) is the same no matter how far or which direction you're looking from.
Mathematically, the monopole moment is the $n=0$ moment, so $$\int{r^n \rho(r) \;d^3(r)} = \int{r^0 \rho(r) \;d^3(r)} = \int{\rho(r) \;d^3(r)} = M$$
The answer to this question gives a good way of thinking about it intuitively.
For mass, there will always be a non-zero monopole moment. But that's not true for other quantities. For example, think about the classic example of a positive charge $q$ and negative charge $-q$ separated by a distance $d$. The total net charge is zero, so the monopole moment is zero, but the dipole moment is not: it is $qd$.
You may have noticed that in this case the dipole moment is independent of any reference point. That's because it's the lowest non-zero moment. In general, moments are a function of a reference point, but the lowest non-zero moment (and only the lowest non-zero moment) is independent of where you set the reference point. So, in the earth example, the monopole moment would be independent of the reference point, but dipole moment (and quadrupole moment, etc.) would not.