Is this the correct way to transform a trajectory between Galilean frames? Consider the general Galilean transformation in one spatial dimension:
$$(x, t) \mapsto (x', t') = (x + vt + a, t + b).\tag{1}$$
I want to use it to transform trajectories $x(t)$ in a frame $S$ to the corresponding trajectories $x'(t')$ in $S'$. Seems simple enough, but when thinking carefully about it I get to unexpected conclusions.
My argument
This is my argument. Note that (1) only explicitly concerns individual coordinate pairs, not functions, so when we try to deduce the form of
$$(x(t), t) \mapsto (x'(t'), t') = \text ?\tag{2}$$
we must be very careful. To remove any ambiguities, consider a particular point in time $t_0$ and the corresponding point $x_0 \equiv x(t_0)$ in space. Then, clearly,
$$(x(t_0), t_0) = (x_0, t_0) \mapsto (x_0', t_0') = (x_0 + vt_0 + a, t_0 + b) = (x(t_0) + vt_0 + a, t_0 + b)\tag{3}$$
so that, with $x'(t_0') \equiv x'_0$, we have
$$x'(t_0') = x(t_0) + vt_0 + a.\tag{4}$$
Now, of course, there was nothing special about the particular time $t_0$, so in general we must have
$$(x(t), t) \mapsto (x'(t'), t') = (x(t) + vt + a, t + b).\tag{5}$$
As for the functional form of $x'(t')$, we see that
$$x'(t') = x(t' - b) + v(t' - b) + a.\tag{6}$$
My doubts
The above seems convincing to me until I consider a pure time translation, $b>0$, $a=v=0$. Then, according to (6), when the clock of an observer in $S'$ reads $t' = T$ they see
$$x'(T) = x(T-b).\tag{7}$$
That is, they see something that occured in the past in $S$. This is the opposite of what I would expect! Shouldn't $S'$ see the future, relative to what $S$ sees?
My question
So where is my mistake? Is my argument wrong, or is it an incorrect assumption that $S$ should "see the future" (for a time translation with $b>0$)? If my argument, and hence equation (5), is wrong, what is the correct argument and equation?
 A: Based upon what OP has said in the comments, let me say that the argument they give appears to be correct as far as I can see. I was initially worried that they ignored the functional dependence of the particle position in the transformation, but this seems not to be the case. Indeed, it appears that OP's concern is sourced from the observation that the argument of $x(t)$ transforms the an inverse fashion to the transformation initially defined (it becomes $T-a$ rather than $T+a$).
This is actually a known "issue," though it's exceptionally rare to see a book actually point this out explicitly. Essentially, whenever we are working with a group and asking how it acts on a function, the argument of the function transforms under the inverse representation of that group. Formally, this can be seen as, more or less, the starting place for invariant theory, which studies how functions transform under the action of a group, with the particular interest in those functions which do not transform (hence, invariant).
Pointing to invariant theory does not, however, give a good argument for why we should believe this as it only points out that some people have found it an interesting proposition to study. I recently came across this absolutely wonderful Math.SE post which I think argues quite nicely for why this fact is natural, though the notation there is maybe a bit abstract if you haven't played much with group actions and representations. So, let me try and convey two of the arguments in that Math.SE post here in terms of translations.
The first thing we could do is just think about the graph of $f(x)=x^2$. If we look at the graph of the function $g(x) = f(x+a) = (x+a)^2$, we would say that the function $g(x)$ is the same as $f(x)$, but with the argument shifted to the right by $a$. However, if we went to plot this function, we would find that the graph of $g(x)$ has actually moved to the left by an amount $a$, not to the right. In order to get a function moved to the right by $a$, we would have had to subtract $a$ from the argument, which is actually moving $a$ to the left.
One way to understand this fact is to say that by writing $x^\prime = x + a$, we are actually moving the coordinate axes to the right by $a$, so of course the graph looks like it has actually moved to the left relative to the new axes. While this makes some amount of sense, I have always found this somewhat unsatisfying because it then begs the question, "but why does this transformation move the axes and not the function?" This explanation, while it makes nice contact with things we have seen before and are used to, fails to provide an answer to this question.
Instead, and this is one of the really good insights in that Math.SE post, we can do something very concrete, which is actually rather similar (but not quite identical to) to the argument OP has made. Let me call the function here $f(t)$ and the coordinates on the plane $(x,t)$ just to avoid potential confusion between talking about the function and its values. The graph of $f$ is then the collection of points $\{(x,t)|x=f(t),t\in\mathbb{R}\}$ in the $(x,t)$ plane.
With this in mind, if we want to think about "translating in time" as very literally taking the collection of points on the curve and moving them to the right in time, then translated graph would be the collection of points $\{(x,t+a)|x=f(t),t\in\mathbb{R}\}$. Importantly, note that we have not changed the condition determining the points, so we still have the $x=f(t)$. We have just taken all such pairs and moved them to the right by $a$. But if we accept this, then since $t-a$ is also a real number, we have
$$
\{(x,t+a)|x=f(t),t\in\mathbb{R}\}=\{(x,t)|x=f(t-a),(t-a)\in\mathbb{R}\} = \{(x,t)|x=f(t-a),t\in\mathbb{R}\}
$$
where we have in the second equality just redefined what we are calling $t$ (since it acts sort of like a dummy variable in the specification of this set anyway).
Looking at this manipulation, we see very explicitly that the function $g(t)$ which produces the collection of points shifted to the right is actually $f(t-a)$. This is a way of looking at this odd feature of function transformations which I hope OP will also find satisfying.
