Why does time-reversal mean negation of time i.e. $t\mapsto -t\;?$ Disclaimer : This is a follow-up to this question.

For long time, I've been pondering of this but couldn't come to a stern conclusion to the question:
Why is time-reversal the negation of time $t\;?$ Meant to say, how does negation of time mean the backward flow of time?
Dumb though the query maybe; but still I want to ask it. 
For instance, this is forward flow of time: $0,10,20,30,40,50\;;$ when it is reversed, the time sequence would look like $50,40,30,20,10,0\;;$ is it the negative of the forward time flow $0,10,20,30,40,50\;?$ No. But still it is the backward flow of time, isn't it?
Why has time-reversal to do with the negation of time, for $50,40,30,20,10,0$ represents the backward flow of time $0,10,20,30,40,50$, although the former is not the negation of the later?
I am sure I'm missing something but couldn't point it out.
So, can anyone explain it to me why time-reversal means $t\mapsto -t\;?$
 A: The transformation $t \mapsto -t$ transforms an ascending function into a decreasing one. It thus inverts the flow of time.
You have to think of it as a transformation on the variable $t$ which is an input in observables : each observable $O(t)$ will see time flowing backwards.
Assume you want to apply the time-inverting transformation at some time $t_0$. Because of time-translation invariance, you can choose the transformation to be $t \mapsto 2 t_0 - t$, it is equivalent to doing just $t \mapsto -t$. You thus get the sequence you described in your question : by the transformation $t \mapsto 100 - t$ starting at $t_0 = 50$, the sequence 50, 40, 30, 20, 10, 0 becomes 50, 40, 30, 20, 10, 0.
A: The flow of time is not simply modelled by the real line. The real line with sum $(\mathbb{R},+)$ is an abelian group that is commonly used to model the time coordinate, and the fact that it is possible to "shift the origin" of time (choose which time should represent the starting point).
The flow of time could be modelled by a function $f:\mathbb{R}\to \mathbb{R}$ that works as follows. Its domain is a time variable, as it is its co-domain (both measured in seconds); and to a time $t\in\mathbb{R}$ we associate the flow $f(t)$ of $t$ seconds by
$$f(t)=t_0 +t\; ,$$
where $t_0$ is a fixed reference time of origin, that is customary chosen to be zero. If $t_0=0$, the function $f$ reduces to the identity function.
Now what does it mean to take the reversed flow $f_\textrm{inv}$, i.e. "reverse time"? It is quite intuitive within the model above: $f_\textrm{inv}:\mathbb{R}\to\mathbb{R}$ such that
$$f_\textrm{inv}(t)=t_0 - t\; .$$
In this way a time lapse of $t$ seconds has flown "backwards" (in the inverse direction wrt the group operation of the coordinates).
Now I would say that the time reversal is the mapping $TR:f\mapsto f_\textrm{inv}$ between the flow and the inverse flow, so it is a map between functions, and not between numbers. If $t_0=0$, it is easy to justify the abuse of notation $TR: t\mapsto -t$. However it remains a map of functions (time flows), and not simply of numbers.
Edit:
[ following the comments ].
To get the two sequences of time the OP mentioned using the flows, we should think as follows:
We start from $t_0$, and look at a sequence of 6 "forward" discrete time steps, each of 10 seconds. Using the flow $f$, we may describe them by $t_0+t$, for any $t\in \{0,10,20,30,40,50\}$. If $t_0=0$, we get the sequence of forward times $0,10,20,30,40,50$. Now we suppose that at the final counted time $t_0'=t_0+50$ we reverse time, and look at a sequence of 6 "reversed" discrete time steps of 10 seconds each. We have now to use the inverse flow $f_{\text{inv}}$, and we get $t'_0-t=t_0+50-t$, for any $t\in\{0,10,20,30,40,50\}$. With $t_0=0$, this gives $50,40,30,20,10,0$.
