Explaining the phrase "as viewed by A, clock $\mathfrak B$ appears to be ticking faster than clock $\mathfrak C$" In writings concerning time dilation and GPS (incl. on PSE) one can find statements such as

When viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground.

To me, this statement and its variants are in need of explanation ...
Apparently it is thereby assumed that the mentioned clocks (for concreteness let's refer to one of the clocks on the satellites as clock $\mathfrak B$, and to one of the clocks on the ground as clock $\mathfrak C$) are each characterized by their particular tick rates; accordingly $\nu_{\mathfrak B}^{\,}$ and $\nu_{\mathfrak C}^{\,}.$
And further, the prescription of those clocks being "identical" is to be understood that all those clocks have equal tick rates, in particular $$\nu_{\mathfrak B}^{\,} = \nu_{\mathfrak C}^{\,}.$$
But then, what exactly is meant by:
"clock $\mathfrak B$ appearing to be ticking faster than clock $\mathfrak C$, as viewed by someone else (say $\mathbf A$)"
??
Would this phrase in fact be referring to a comparison of certain rates of $\mathbf A$, namely $\mathbf A$'s rate of receiving the tick signals issued by clock $\mathfrak B$ being greater than $\mathbf A$'s rate of receiving the tick signals issued by clock $\mathfrak C$; symbolically:
$$\nu_A^{(\circledR \, \mathfrak B)} > \nu_A^{(\circledR \, \mathfrak C)}$$
?
 A: The quoted sentence is probably intended to mean what you suggest: the two clocks emit ticks, which are received somewhere, and the ratio between the time-averaged rates at which the ticks are received isn't $1$.
This omits details about the nature of the tick signals (light, sound, etc), and where the receiver (your $A$) is located, but the ratio of the rates turns out to be independent of those details, as long as the whole system including the signals and receiver is in a quasi-steady state. (That condition is intended to rule out silly situations like bouncing the signal from the satellite off of an ever-receding mirror, or using slower-than-light signals that get slower and slower as time goes on.)
A: The statements are misleading at best and possibly nonsense. In SR all (ideal) clocks measure time at the same rate in their respective rest frames. The phenomenon referred to as time dilation arises because the planes of simultaneity of two references frames in motion relative to each other are tilted, so a level plane in one frame, across which it is the same time everywhere, is a tilted slice through time in the other frame, and vice versa.
To take a concrete example, suppose you start walking down a corridor at 1m/s beginning at time t=0. Upon the wall where you start is a clock identical to your own, also showing t'=0. Along the corridor at 5m intervals are other clocks identical to you own and ticking at exactly the same rate as your own. However, each of the clocks placed along the corridor has been set to be 1 second ahead of its nearer neighbour.
After you have walked for 5 seconds you reach the first clock along. Your clock reads t=5, but the clock on the wall, having been set 1 second ahead, reads t'=6. When you reach the next clock 5 seconds later, your reads t=10s but the adjacent clock being two seconds ahead reads t'=12s. At the next clock, yours reads t=15s and it reads t'=18s. And so on down the corridor. All the clocks- yours and those one the wall- are ticking at exactly the same rate, but the cumulative time on your clock gets progressively further behind the time shown on the walls clocks as you move down the corridor- in other words, your clock seems to running slow. It is doing nothing of the sort, it is simply that the wall clocks were already reading progressively later times at the moment you set off.
Exactly the same effect is the cause of relativistic time dilation. When you start to move relative to another frame at t=0, it is t=0 everywhere in your frame. In the other frame, however, it is only t'=0 where you are- further ahead in your direction of travel  it is already a progressively later time. Your clock and all the clocks in the other frame are measuring time at the same rate, but since you are moving between successive clocks in the other frame, each of which started off ahead of the previous one, the time shown on your clock is progressively less than the readings on the clocks you pass.
So you will see that a statement that the moving clock runs slower than the stationary one is wrong, and shows a complete misunderstanding of what is happening. It is also a source of endless confusion and questions about 'how can both clocks be running slower than the other?'
A: The main principle in relativity theory is that anyone will see physical processes in their local vicinity proceeding like normal. Any time you have to observe something far away, the rate of those processes may be different from the far away perspective, because of relative motion or gravitational fields (which are a type of relative motion).
The far away observation could be accomplished in any number of ways, e.g. through a telescope, or through the far away clock sending regular signals that the other observer receives.  Does that help?
