In day, when you look in the room through the window out, you can clearly see what happens outside. At night when it's dark outside but there's light inside you can look in the window but it becomes a mirror.
Why?
Optics – Why Does a Window Become a Mirror at Night?
opticsreflectionvisible-light
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You were right with your original though. This is an effect purely due to a lack of convective cooling. When you are inside, there is normally very little to no airflow, which means when you stand in the sunlight and heat up due to it, the air around you will be warmed by you but isn't moving enough to be replaced with cool air and carry the heat away. Don't misunderstand, the air is moving, which is why you won't continue to build up more and more heat, your temperature will plateau. It's just that the air isn't moving enough to prevent an increase in skin temperature. When you step out of the direct sunlight, it feels cool again because now you are not being heated as much and what convective flow exists is enough to cool you to normal levels. Additionally, there is a component that is attributed to the air being directly heated by the sunlight as well as re-radiation from heated objects (like a table or chair), but most is due to a lack of convective cooling.
As for your anecdotal evidence, I cannot comment on why it might have felt just as hot with a strong fan on you, but I can try to convince you that my answer is true. If you go outside one morning when there is little to no wind/airflow and stand in the sunlight, it will feel just as hot. Even in the winter, the sunlight will feel hot, however the air around you will have slightly more cooling power so it may still feel cold.
You are getting reflections from the front (glass surface) and back (mirrored) surface, including (multiple) internal reflections:
It should be obvious from this diagram that the spots will be further apart as you move to a more glancing angle of incidence. Depending on the polarization of the laser pointer, there is an angle (the Brewster angle) where you can make the front (glass) surface reflection disappear completely. This takes some experimenting.
The exact details of the intensity as a function of angle of incidence are described by the Fresnel Equations. From that Wikipedia article, here is a diagram showing how the intensity of the (front) reflection changes with angle of incidence and polarization:
This effect is independent of wavelength (except inasmuch as the refractive index is a weak function of wavelength... So different colors of light will have a slightly different Brewster angle); the only way in which laser light is different from "ordinary" light in this case is the fact that laser light is typically linearly polarized, so that the reflection coefficient for a particular angle can be changed simply by rotating the laser pointer.
As Rainer P pointed out in a comment, if there is a coefficient of reflection $c$ at the front face, then $(1-c)$ of the intensity makes it to the back; and if the coefficient of reflection at the inside of the glass/air interface is $r$, then the successive reflected beams will have intensities that decrease geometrically:
$$c, (1-c)(1-r), (1-c)(1-r)r, (1-c)(1-r)r^2, (1-c)(1-r)r^3, ...$$
Of course the reciprocity theorem tells us that when we reverse the direction of a beam, we get the same reflectivity, so $r=c$ . This means the above can be simplified; but I left it in this form to show better what interactions the rays undergo. The above also assumes perfect reflection at the silvered (back) face: it should be easy to see how you could add that term...
Best Answer
Any translucent surface both reflects and refracts light. By refraction, I mean that it bends the light a bit, but lets it through to the other side. Now, reflection for such surfaces is much less than refraction (unless there's total internal reflection, but thats irrelevant for glass+air). Edit: According to @JohnRennie (see comments), only 5% of the light is reflected
During the day, you have light from your room being largely refracted out, and reflected back inwards a tiny bit. The outside light does someing similar. It is largely refracted into your room, and reflected back outside a tiny bit. So, the majority of the light you see coming from the window is due to the outside light. You will see a reflection if you look carefully (exacly how carefully depends upon the lighting of your room)
Now, during the night, there is little or no light coming from the outside. So the majority/all of the light you see is due to reflection. So you see the reflected image.
Now an interesting question is, if the reflected image has the same intensity in both cases, why do you see it in one case and not see it in another? The answer lies in the working of the eye. The eye does not have a constant sensitivity to light. Whenever there is a lot of light, your irises contract, admitting less light into your eyes. This means that you can perceive bright light but dim light becomes invisible. When it is darker, they expand, and the reverse effect happens. That's why you feel blinded by bright light when you leave a dark room, and also why it takes time to adjust to a dark room. (You can actually see your irises contracting; go to a well lit room with a mirror, stare at your eyes, close them for a few seconds, then reopen.. Takes a few tries, but you can see them contracting). Edit: (Credit @BenjaminFranz for pointing this out) The regulatory mechanism does not consist of only the iris/pupil. The retina also does a lot of regulation, which is why it takes half a minute or more to get used to a dark room, whereas our irises can dilate within a few seconds.
So, during the day, the profusion of light refracted from the outside makes your irises contract, thus making the reflected light nearly invisible. During the night, your pupil is dilated, so you can clearly see a reflection.