I think you are overthinking this?
The basic physical fact that we are trying to reconcile in special relativity is that when you accelerate in some direction $\hat z$ with acceleration $a$, you see clocks ahead of you tick faster, and clocks behind you tick slower, at a rate $1 + az/c^2$. This includes a surface at $z=-c^2/a$ where clocks appear to “stand still.” After a certain period of acceleration, it will appear to you that all of these clocks that appeared to be in-sync when they were at rest relative to you, now no longer appear to be in-sync. Your notion of the “present moment” is therefore different, at long distances, from other folks who are moving relative to you, by an amount roughly equal to $vz/c^2$.
Things like time dilation and length contraction are indirect consequences of this central physical effect as it interacts with the standard effect of acceleration which everyone knows about (that things start moving in the $-\hat z$ direction).
In the twin paradox, one twin leaves Earth and then accelerates back towards Earth; this causes them to see Earth’s clock tick very very fast as they accelerate; their hyperplane of “the present moment” is tilting about them and the twin on Earth suddenly goes from aging slowly (due to time dilation) to aging quickly (due to the above effect), and then once acceleration is complete they return to aging slowly. In your modification of the twin paradox, one twin leaves Earth and then shares information with someone whose present-hyperplane is already tilted appropriately as that twin would have had; the two communicators simply disagree on when the present is all the way over at Earth. Both of these resolve the paradox the exact same way; one of them just hides the acceleration in an information-propagation step—but it cannot stop the present-hyperplane of the incoming spaceship from being tilted in precisely that way. This is not an “everybody gets limitless freedom to choose the present to be whatever they wish” situation; accelerations are the only way to twist this present-hyperplane and they are consistent; any combination of accelerations which results in the same ending velocity results in the same time-direction in spacetime and hence the same present-hyperplanes normal to that time-direction.
In fact, and this is somewhat of a subtle subject, everyone agrees on the proper time of a worldline segment. Everyone can calculate it, everyone knows that this is the time measured by an observer on that worldline. Everyone can calculate that there is no difference between the age of a twin who departs Earth and accelerates back towards it, and the age of some information being transported to the same turn-around spot and then being transferred to another spaceship which then races back towards Earth. None of the invalid operations are happening on the traveling twin's worldline; that twin ages slow and that is very obvious. The only things that you are doing in the twin paradox which are invalid come from reasoning about what the traveling twin sees about the twin who remained on Earth, since the traveling twin saw that twin age slowly across the portions-in-motion of their journey but, at turning around, must have seen the twin suddenly jump in age from one amount to another amount which cancels out all of this slow aging. (Well, the above is not quite right; the Earth-twin does appear to be aging faster than normal while the space-twin is approaching them and vice versa; but the point is that it is not as fast as they should be aging after considering the Doppler effect.)
Your rule 1 is incorrect therefore—you can compare worldline segments just fine even if they do not have the same endpoints. However it is much harder to generate an obvious contradiction or paradox with this comparison, because the relativistic present has some flexibility; any spacelike hyperplane is a potential relativistic present for someone lying on it with suitable motion. So if they are not at the same place it is possible that different observers say “this happened first, then that” in different ways.
Similarly rules 2 and 3 are problematic; you can say just fine that the traveling twin on their outbound leg ages by some $\Delta t/\gamma$; there is nothing wrong with that. You can say that in the Earth-twin's present this corresponds to them having aged $\Delta t$, that is also fine. But since these are vastly space-separated you do need to be sensitive to the fact that the outbound twin, pre-acceleration, draws the present differently and thinks the present moment for the Earth twin is $\Delta t~\left(1 - \frac{v^2}{c^2}\right).$ And then as they accelerate towards the Earth twin this jumps suddenly to being $\Delta t \left(1 + \frac{v^2}{c^2}\right)$ so that after another $\Delta t/\gamma^2$ they appear to have aged by $2\Delta t$ when they both rendezvous again, compared to the traveling twin's $2\Delta t/\gamma.$
Appendix: how this leads to the Lorentz transformation
In relativity we define $w = ct$ to be a measure of time with the units of space, and then our acceleration is some parameter $\alpha = a/c^2$ in units of inverse distance. Then we can write the transformation given above, as saying that if you accelerate for a time $\Delta w$ (where $\Delta$ is just a symbol meaning “a change in”), your coordinates $(w,x,y,z)$ transform like$$\begin{align}
w' &= w - \alpha ~\Delta w~z,\\
x' &= x,\\
y' &= y,\\
z' &= z - \alpha ~\Delta w~w.\end{align}$$
These are easier to handle if you replace $(w, z)$ by $p = w + z$ and $q = w - z$ (and later you can recover $w = (p + q)/2, z = (p-q)/2$), to find
$$\begin{align}
p' &= (1 - \alpha~\Delta w) p,\\
q' &= (1 + \alpha~\Delta w) q,\\
x' &= x,\\
y' &= y.\end{align}$$
So in these coordinates $p'$ gets multiplied by a number, and q gets multiplied by a different number.
Now the problem is that this only works well if $\Delta w$ is small, it breaks if $\Delta w$ is big. So for example if $\alpha~\Delta w = 1$ then we get $p'=0$ and hence $w' = -z'$ for our coordinates, which does not sound good. What is actually creating this problem is, as I am accelerating, my coordinates $(w, z)$ are changing, so I am supposed to actually, for longer accelerations, divide the total time of acceleration $W$ into $N$ chunks for some large $N$ and then repeat this procedure over and over with the new coordinates. As a result $p'$ will be repeatedly multiplied by one number and $q'$ will be repeatedly multiplied by a different number.
Now there happens to be a special number $e\approx 2.718281828\dots$ such that $e^x \approx 1 + x$ for very small $x$; you don't need to know exactly how we get this number I don't think, but it exists and if $N$ is large enough we can assume $1 + \alpha W/N \approx e^{\alpha W/N}$ to very high precision. The nice thing about converting this to an exponent in this way is that repeated multiplication becomes repeated addition in the exponent, so that after $N$ iterations you will find
$$\begin{align}
p' &= \left[e^{-\alpha W/N}\right]^N~p,\\
q' &= \left[e^{+\alpha W/N}\right]^N~q.\end{align}$$
and the $N$s cancel. So the general rule must be, for some $\phi = \alpha W$, that $$(w' + z') = e^{-\phi}~(w + z),\\
(w' - z') = e^{+\phi}~(w - z),$$
which we can solve directly as $$ w' = \frac{e^{\phi} + e^{-\phi}}{2} w - \frac{e^{\phi} - e^{-\phi}}{2} z\\
z' = \frac{e^{\phi} + e^{-\phi}}{2} z - \frac{e^{\phi} - e^{-\phi}}{2} w,$$
which now holds for all $\phi$. This is the full Lorentz transformation, it includes time dilation and length contraction effects due to this term $\gamma = (e^{\phi} + e^{-\phi})/2,$ which was not in the above expression.
Now to connect these exponents $\phi$ to an actual speed, it would be helpful to find a trajectory $z = v~t$ which is now stationary, $z'=0$. To do this we invent the dimensionless velocity $\beta = v/c$ and substitute $z = \beta~w$ into the second equation and then set it equal to zero to find the $\beta$ that makes $z'=0$, and this turns out to be $$\beta = \frac{e^\phi - e^{-\phi}}{e^\phi + e^{-\phi}}.$$ Looking above, we can rewrite $(e^{\phi}-e^{-\phi})/2$ as simply $\gamma\beta,$ So we can use our $\gamma, \beta$ to rewrite the above as $$ \begin{align} w' &= \gamma ~w - \gamma\beta~ z,\\
x' &= x,\\
y' &= y,\\
z' &= \gamma~ z - \gamma\beta~ w.\end{align}$$
This is the format that we usually present the Lorentz transform to students for the first time.
Finally, you should be very suspicious about having one free parameter $\phi$ and then suddenly having two, $\gamma,\beta.$ Indeed they are dependent on each other, and the connection can be made apparent by simply squaring $$\left(\frac{e^\phi \pm e^{-\phi}}{2}\right)^2 = \frac{e^{2\phi}}4 \pm \frac12 + \frac{e^{-2\phi}}4,$$ so that the squared-$+$-case is one plus the squared-$-$-case, as they agree on the two terms on the outside. As a direct result, $$\gamma^2 = 1 + \gamma^2 \beta^2\\
(1 - \beta^2) \gamma^2 = 1\\
\gamma = \frac1{\sqrt{1 - \beta^2}}.$$This is the more familiar way that $\gamma$ is usually defined, but the $\phi$-version is very helpful: we call these functions of $\phi$ the “hyperbolic sine/cosine” and $\phi$ itself is the “rapidity” corresponding to velocity $\beta$; it turns out that special relativity has a rapidity-addition formula rather than a velocity-addition formula; velocities add only in this rapidity-space.
