I've also never found a text which explains any of this particularly well, so I'll try to give my own explanations after years of piecing bits together. I've come across four common magnification definitions and I give a brief description of each. There is a lot of text but I hope it helps.
Linear magnification
Even though you're asking about angular magnification, I'll include this for completeness. In this case the system produces an image in the image plane of an object in an object plane, both of which are not at infinity (note also that the image can be either real or virtual). The linear magnification is then simply defined as the linear size of the image (i.e. its physical length in the image plane) to the linear size of the object.
Here there is no need to talk about an eye anywhere.
Angular magnification of a telescope
Now the object is at infinity: the rays from any given point on it can be assumed to be parallel when they reach the entrance pupil of the system. However, the rays from a point on one side of the object will enter at a different angle (to the optic axis) to the rays from a point on the other side. This is where angular size comes in, defined as the angle subtended by the object at the entrance pupil, or equivalently the difference in angle of the rays from opposite sides of the object.
When such parallel rays from one side of the object pass through the system and leave the exit pupil, they will again be parallel (telescopes are designed to form an image at infinity of an object also at infinity), but at a different angle to the optic axis. Similarly, rays from the other side will be also be parallel and at a different angle. The angular size of the image is now defined as the difference between these new angles. Magnification is then simply the ratio of the angular size of the image to the angular size of the object.
If you want to discuss an eye here, it can technically be anywhere after the exit pupil as the rays are all parallel and you will see the same thing at any distance away. However, of course all the rays must actually enter your eye, so it is sensible to place your eye close to the exit pupil.
I hope this answers your question about the moon, something which would only ever be imaged with a telescope. The rays are not focussed onto a focal point but rather leave parallel as discussed above, and there is no need to discuss the near-point etc.
Angular magnification of a microscope
As in the case of linear magnification, the object is in an object plane which is not at infinity. However, the image is now formed at infinity. In the same way as for a telescope, rays from a given point on the object leave the exit pupil parallel and at a certain angle to the optic axis. The angular size of the image is defined as the difference between the angles of the rays leaving the exit pupil which originate from opposite sides of the object.
However, this time it doesn't make sense to talk about the angular size of the object from the point of view of the entrance pupil (which is generally very close to the object, compared to the telescope where it is at infinity), as a human eye would never be able to see it so close up. Rather, the best view the unaided human eye can get is when it is placed at the near point, so for a 'fair' comparison we work out the angular size of the object when it is placed at the near point with no imaging system. This is defined in terms of the tangent (which assumes one end of the object is on the optic axis, a convention many books use): if $h$ is the linear size of the object and $D$ the near point distance then the angular size of the object is $\alpha=\arctan(h/D)$ (but generally angles are small enough that $\alpha\approx h/D$ anyway). The magnification is now defined as the ratio of the angular size of the image to the angular size of the object at the near point.
Thus your example calculation for the "size of the object unaided" is correct, but your calculation for the angular size of the image is wrong; for that you'd need to know the specifics of the microscope to work out how points on the object are mapped to angles in the image.
Again, the eye can be anywhere after the exit pupil, but sensibly somewhere close such that all the rays actually go in.
Angular magnification of a magnifying glass
In my opinion this is the most annoying one. I'll consider the case where the magnifying glass is pressed right up to the eye (which is not how anyone ever uses one but seems to be the case covered by many optics books).
As with linear magnification, neither the object nor the image are at infinity, but (as long as the object is placed within the focal length of the magnifying glass) the image is always virtual. As the eye is placed right up to the glass, it simply sees a virtual image of linear size $H$ a distance $v$ away, the angular size of which is given by $\beta=\arctan(H/v)\approx H/v$. Just like for the microscope, for a fair comparison, we define the angular size of the object as that when it is at the near point with no imaging system, and the magnification is given by the ratio of the angular size of the image to the object as usual.
Just a final note. I've made a clear distinction between things being at infinity or not. However, something at infinity can still have its angular size defined in terms of linear size. It's just that its linear size is also infinite (of course in reality it's not but as far as the calculation is concerned) and so it is unhelpful to talk about it in terms of this.
Best Answer
A telescope with two convex (converging) lenses is a Keplerian telescope. The lens with the longer focal length is the objective, and the shorter focal length lens is the eyepiece. Since it is explicitly stated that the lenses are thin, you can use the thin lens equations:
$$ \frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f} $$
where $d_i$ is the distance to the image, $d_o$ is the distance to the object, and $f$ is the focal length. You will also need
$$ M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} $$
where $M$ is the magnification, $h_i$ is the image height, and $h_o$ is the object height. Note the minus sign: you need to follow sign conventions on the image and object distances -- see any introductory physics textbook for coverage on these. A negative magnification represents an inverted image.
Whenever you encounter a problem like this, it is always best to draw a ray diagram. Consider the diagrams below:
The first diagram shows the typical situation, where the intermediate image is outside of the focal length of the second lens. The second diagram shows the situation you are interested in, where the image of the first lens falls inside the focal length of the second lens. This yields a virtual image between the two lenses.
Using the first equation, we calculate $d_{i1}$: $$ d_{i1} = \frac{d_o f_1}{d_o - f_1} = 105.263 \ \text{cm} $$ Now, this yeilds a magnification for the first image of $$ M_1 = -\frac{d_{i1}}{d_o} = -0.0526 $$
As you have already found, the separation between the lenses is $L=112.406 \ \text{cm}$ and you are given that $d_{i2} = -25 \ \text{cm}$. The minus sign tells you the image is formed on the "wrong side" (left in the image) of the lens, i.e., it is a virtual image. This implies that $$d_{o2} = L - d_{i1} = 7.143 \ \text{cm}$$
Next, we compute the second magnification: $$ M_2 = -\frac{d_{i2}}{d_{o2}} = 3.500 $$
Now, the total magnification is simply the product of the two:
$$ M_{total} = M_1 M_2 = -0.184 $$
However, what you really want is the angular magnification.
Assume, for simplicity, that the original object height is $1 \ \text{cm}$. That means that the final virtual image will be $0.184 \ \text{cm}$ tall and inverted. However, the original image is $d_o + L = 2112.406 \ \text{cm}$ away, which implies an angular size (using small angle approximation) of: \begin{equation} s = \theta r \implies \theta_1 = \frac{s}{r} = \frac{1 \ \text{cm}}{2112.406 \ \text{cm}} = 4.734 \times 10^{-4} \ \text{radians} \end{equation}
The virtual image, on the other hand, will have an angular size of \begin{equation} \theta_2 = \frac{s}{r} = \frac{0.184 \ \text{cm}}{25 \ \text{cm}} = 7.36 \times 10^{-3} \ \text{radians} \end{equation}
Thus the angular magnification will be \begin{equation} M_{angular} = \frac{\theta_2}{\theta_1} = \frac{7.36 \times 10^{-3} \ \text{radians}}{4.734 \times 10^{-4} \ \text{radians}} = 15.547 \end{equation} I have been sloppy with rounding off my numbers, but this should give you the idea. Cheers!