There are lots of questions about the twin paradox on this site, so it's probably not worth going over that material again.
What is worth saying is that where people tend to get confused is by misunderstanding what an inertial frame is and how different inertial frames can be compared. We should simplify matters a bit and put twin B on a spaceship because orbital motion is a bit more complicated. The only time A and B can directly compare anything with each other is the moment that they pass i.e. the moment that they are in the same place. If A and B stay in their inertial frames they will never meet again and indeed will get further and further apart as time passes. The only way the twins will ever meet again is if they change inertial frames i.e. if one of them accelerates.
In SR acceleration is absolute. By this I mean that velocity is relative i.e. you cannot tell whether A or B is the one moving, but it is always possible to tell which of the two is accelerating. The acceleration always introduces an assymetry so it's no surprise that when they meet again A and B will find their clocks differ.
You can treat acceleration in special relativity. See for example my answer to How do I adjust the kinematic equations to avoid reaching speeds faster than light? where I give (some of) the equations for understanding the motion of an accelerating rocket. If you do the calculation you will find that B sees A clock running fast while B is accelerating between inertial frames. See my answer to Why isn't the symmetric twin paradox a paradox? for more on this.
The question of what Carl Sagan's neutrino sees is quite a subtle one. Suppose some particle interaction shortly after the Big Bang and 13.7 billion years away produced a neutrino and that neutrino has just passed you. For the neutrino only a few seconds has passed since the Big Bang. However when the neutrino passes you it sees you at your current age, 13.7 billion years, so what's going on? The answer is that in the neutrino's frame and your frame the Big Bang happened at different times. So the neutrino can see the 13.7 billion age of the Big Bang pass in a few seconds, but not because it sees the universe's time running fast. It sees each successive bit of the universe as older because in each bit of the universe it passes through the Big Bang gets further and further back in time.
You are right that there are two effects at play here.
Firstly, suppose that we turn relativity off --- suppose that we consider our universe to be Newtonian, with a finite speed of light propagation. It would indeed be the case that if an observer A were moving relative to another observer B, and emitting a light signal towards B, then the rate at which the signal arrived at B would be different to that at which it was emitted. This is the plain old Doppler effect, just the same as the effect experienced when listening to an ambulance moving towards or away from you. If the signal emitted by A was the light from her clock face, then indeed observer B would see the clock tick at a different rate.
But would she see it tick more quickly or more slowly? Now that depends on whether observer A were moving towards or away from observer B. If A were moving towards B, B would see the clock tick more quickly, just as you hear the ambulance siren at a higher pitch if it is moving towards you. Likewise, if A were moving away from B, B would see the clock tick more slowly, just as you hear the ambulance siren at a lower pitch if it is moving away from you. Of course, for speeds small compared to the speed of light, this effect would be very small.
Now suppose we turn relativity on. A new effect comes into play --- one much more profound and weird. The (relativistic) phenomenon of time dilation would entail that, according to B, the clock belonging to observer A runs intrinsically more slowly. It's not that she merely sees the the clock tick more slowly, it's that time itself has slowed down aboard A's spaceship, according to B. What this means is that if B were to measure the rate of arrival of signals from A and then do some maths to undo the Doppler effect, such that the resulting quantity obtained were the rate of emission of the signal, she would find the rate to be less than that at which A would say she emitted the signal. Which is pretty nutty.
So what is the combined effect? Well, we can see immediately that they won't cancel out in all cases by considering observer A to be moving away from observer B. Then time dilation and the Doppler effect combine, such that observer B sees the clock aboard A's ship running particularly slowly. The correct formula (I won't give a derivation here, but you can find many on the web, e.g. on Wikipedia) turns out, in fact, to be
$$ f_O = f_E \sqrt{\frac{1-\beta}{1+\beta}} \,,\qquad \beta = v/c\,,$$
where $f_E$ is the emitted frequency and $f_O$ the observed frequency. For all positive velocities, which in this case corresponds to observer A moving away from observer B, the top of the fraction is smaller than the bottom, and so the observed frequency is less than the emitted frequency. Likewise, for all negative velocities, which in this case corresponds to observer A moving towards observer B, the top of the fraction is larger than the bottom, and so the observed frequency is greater than the emitted frequency.
Qualitatively, then, the two effects combined are no different from the Doppler effect by itself. The only circumstance under which the two effects 'cancel out' is when the observers' relative velocity is zero, and in that case the two effects are individually zero! Whilst the Doppler effect results in an increased rate for $\beta < 0$ and a decreased rate for $\beta >0$, the phenomenon of time dilation always acts to slow the rate down. Then it just happens to be the case that for no choice of relative velocity can time dilation slow the rate down quite enough to counteract an increased rate due to the Doppler effect.
I've plotted the square root in the above formula below:
You can hopefully see that this function is always greater than 1 for $\beta$ less than zero, and always less than 1 for $\beta$ greater than zero. Hope this helps!
Best Answer
It's reconciled by the relativity of simultaneity. Each observer is assumed to measure the time of events using local readings on a network of clocks at rest relative to that observer, which are defined as "synchronized" in that observer's own rest frame using the Einstein synchronization convention. But because this convention is based on each observer assuming that light moves at a constant speed relative to themselves, different observers will disagree about simultaneity, each thinking the clocks of the other are out-of-sync.
So, if one of your clocks is moving at 0.6c relative to me, I might first measure your clock reading T=0 seconds as it passed a clock of mine also reading t=0 s, then later I might measure your clock reading T=20 s as it passed a clock of mine reading t=25 s, at a distance of 15 light-seconds to the right of my other clock according to my ruler. From this I conclude that your clock only ticked 20 seconds in 25 seconds of my time, meaning your clock was slowed down by a factor of 0.8--but this depends on the assumption that my two clocks were synchronized, which isn't true in your own frame. If two clocks have a distance of L between them in their rest frame and are synchronized in that frame, then from the perspective of another frame which sees them moving at speed v, at any given moment the clock in the rear will have a time that's ahead of the clock in the lead by an amount $vL/c^2$, so in this example where the two clocks are 15 light-seconds apart in my rest frame, and you see them moving at 0.6c, the rear clock will have a time that's ahead of the lead clock's time by (0.6 light-seconds/second)(15 light-seconds)/(1 light-second/second)^2 = 9 seconds. So in your frame, at the same moment your clock was reading T=0 s and passing my left clock reading t=0 s, my right clock (which is in the 'rear' from your point of view, since both are traveling left in your frame) read t=9 seconds. Then 20 seconds later in your frame, your clock passes my right clock, and in your frame it's my clocks that are running slow by a factor of 0.8, so after 20 seconds my clocks have only ticked forward by 16 seconds. But since the right clock started at t=9 seconds, then after ticking an additional 16 seconds it will have reached t=25 seconds. So this is how you can consistently explain the fact that when your clock passed my right clock, your clock read T=20 s and mine read t=25 s, even though my clocks were running slower than yours in your own frame.
Here are some diagrams showing in a more visual way how observers using their own rulers and clocks can each measure the other's rulers as length-contracted and the other's clocks as slowed-down and out-of-sync, in a completely symmetrical way. In this example, we have two rulers with clocks mounted on them moving alongside each other, and in order to make the math work out neatly, the relative velocity of the two rulers is (square root of 3)/2 * light speed, or about 259.628 meters per microsecond. This means that each ruler will observe the other one’s clocks tick exactly half as fast as their own, and will see the other ruler's distance-markings to be squashed by a factor of two. Also, I have drawn the markings on the rulers at intervals of 173.085 meters apart—the reason for this is again just to make things work out neatly, it will mean that observers on each ruler will see the other ruler moving at 1.5 markings/microsecond relative to themselves, and that an observer on one ruler will see clocks on the other ruler that are this distance apart (as measured by his own ruler) to be out-of-sync by exactly 1 microsecond, some more nice round numbers.
Given all this, here is how the situation would look at 0 microseconds, 1 microsecond, and 2 microseconds, in the frame of ruler A:
And here’s how the situation would look at 0 microseconds, 1 microsecond, and 2 microseconds, in the frame of ruler B:
Some things to notice in these diagrams:
in each ruler's frame, it is at rest while the other ruler is moving sideways at 259.6 meters/microsecond (ruler A sees ruler B moving to the right, while ruler B sees ruler A moving to the left).
In each ruler's frame, its own clocks are all synchronized, but the other ruler's clocks are all out-of-sync.
In each ruler's frame, each individual clock on the other ruler ticks at half the normal rate. For example, in the diagram of ruler A’s frame, look at the clock with the green hand on the -519.3 meter mark on ruler B--this clock first reads 1.5 microseconds, then 2 microseconds, then 2.5 microseconds. Likewise, in the diagram of ruler B’s frame, look at the clock with the green hand on the 519.3 meter mark on ruler A—this clock also goes from 1.5 microseconds to 2 microseconds to 2.5 microseconds.
Despite these differences, they always agree on which events on their own ruler coincide in time and location with which events on the other. If you have a particular clock at a particular location on one ruler showing a particular time, then if you look at the clock right next to it on the other ruler at that moment, you will get the same answer to what that other clock reads and what marking it’s on regardless of which frame you’re using. Here’s one example:
You can look at the diagrams to see other examples, or extrapolate them further in a given direction then I actually drew them, and to later times, to find even more. In every case, two events which coincide in one reference frame also coincide in the other.