I find two errors/mistakes in this part:
First, although the definition of $$ \tilde{\gamma}$$ is right the expression you obtained is wrong
How so? The easiest way to work it out is to calculate $$\tilde{\gamma}^{2}$$ and then take the square root of the final result:
$$
\begin{eqnarray}
\tilde{\gamma}^{2} &=& \frac{1}{1 \,-\, \tilde{v}^{2}/c^{2}} \quad \text{ with } \tilde{v} \,=\, \frac{2v}{1 \,+\, v^{2}/c^{2}} \\
&=& \frac{1}{1 \,-\, \frac{4 v^{2} / c^{2}}{(1 \,+\, v^{2}/c^{2})^{2}}} \\
&=& \frac{(1 \,+\, v^{2}/c^{2})^{2}}{(1 \,+\, v^{2}/c^{2})^{2} \,-\, 4 v^{2}/c^{2}} \\
&=& \frac{(1 \,+\, v^{2}/c^{2})^{2}}{(1 \,-\, v^{2}/c^{2})^{2}} \,,
\end{eqnarray}
$$
So $$\tilde{\gamma} \,=\, \frac{1 \,+\, v^{2}/c^{2}}{1 \,-\, v^{2}/c^{2}}$$ like I said.
Second, the set of the two equations of the Lorentz transform you developed for the right-twin:
$$
\begin{eqnarray}
t' &=& \tilde{\gamma} \tau_{\mathrm{R}} \,-\, 2 \gamma \frac{v}{c^{2}} D \\
x' &=& 2 \gamma D \,-\, 2 \gamma^{2} v \tau_{\mathrm{R}} \,,
\end{eqnarray}
$$
presents two different gammas and so seems like you are "mixing frames" someway. I agree the right twin has associated a $$ \tilde{\gamma}$$ but only this one must appear in the equations.
I'm not sure this is clear to you, but the Lorentz transformation I applied is from the "rest" frame (where I defined the scenario in a symmetric way) to the left twin's frame (because I want the left twin's point of view), so the transform's velocity is $$v$$ and the Lorentz factor is $$\gamma = \gamma(v)$$ (without the tilde). The $$\tilde{\gamma}$$ just comes from applying that Lorentz transformation to the right twin's trajectory and simplifying a bit (you get a factor of $$\gamma^{2} (1 \,+\, v^{2}/c^{2})$$, which as I explained above is simply $$\tilde{\gamma}$$).
I didn't bother developing much further since I didn't find it necessary, but it's also not difficult to show that $$2 \gamma^{2} v \,=\, \tilde{\gamma} \tilde{v}$$, which you can use to write the equation for $$x'$$ as
$$x' \,=\, 2 \gamma D \,-\, \tilde{\gamma} \tilde{v} \tau_{\mathrm{R}} \,.$$
If you rearrange the expression for $$t'$$ and substitute, you can write the right twin's trajectory in the left twin's frame as
$$x' \,=\, 2 \gamma D (1 \,-\, \frac{\tilde{v}v}{c^{2}}) \,-\, \tilde{v} t' \quad \text{ (for } t' \,>\, 0 \text{)} \,, \quad(*)$$
so we find that the right twin's velocity is $$-\tilde{v}$$, as we should, and $$2 \gamma D (1 \,-\, \frac{\tilde{v}v}{c^{2}})$$ just represents the right twin's initial position in the left twin's frame just after he's accelerated, given that they start a distance $$2 D$$ apart in the "rest" frame (and each twin's own frame just before accelerating).
What the left twin observes can be deduced by a Lorentz transform with this $$ \tilde{\gamma}$$ value in the equations and the first $$ \gamma$$ must not appear. It must work independently of the observation and so the transform and its gamma value of the not travelling twin.
I stated this above but it's important so I'll insist on it. I defined the scenario in the "rest" frame, and I was applying a Lorentz transformation from the
rest frame to the left twin's frame, and not from the right twin's frame. I never just assumed I should use $$\tilde{\gamma}$$ and $$\tilde{v}$$. I always used $$\gamma$$ and $$v$$ and then recovered $$\tilde{\gamma}$$ and $$\tilde{v}$$ in some places just by simplifying.
For example, you get the expression for $$t'$$ by substituting $$t \,=\, \gamma \tau_{\mathrm{R}}$$ and $$x \,=\, 2D \,-\, \gamma v \tau_{\mathrm{R}}$$ into $$t' \,=\, \gamma(t \,-\, \frac{v}{c^{2}} x)$$. Developing this you get:
$$
\begin{eqnarray}
t' &=& \gamma \bigl( \gamma \tau_{\mathrm{R}} \,-\, \frac{v}{c^{2}} ( 2D \,-\, \gamma v \tau_{\mathrm{R}} ) \bigr) \\
&=& \gamma^{2} \tau_{\mathrm{R}} \,-\, 2 \gamma \frac{v}{c^{2}} D \,+\, \gamma^{2} \frac{v^{2}}{c^{2}} \tau_{\mathrm{R}} \\
&=& \gamma^{2} (1 \,+\, v^{2}/c^{2}) \tau_{\mathrm{R}} \,-\, 2 \gamma \frac{v}{c^{2}} D \\
&=& \tilde{\gamma} \tau_{\mathrm{R}} \,-\, 2 \gamma \frac{v}{c^{2}} D \,.
\end{eqnarray}
$$
(Notice something important here: I assumed the right twin had a velocity $$-v$$ and a time dilation factor of $$\gamma$$ in the "rest" frame, and from that I am
deriving that his velocity is $$-\tilde{v}$$ and time dilation factor is $$\tilde{\gamma}$$ in the left twin's frame after the acceleration. Assuming one implies the other.)
Similarly, if you substitute $$t \,=\, \gamma \tau_{\mathrm{L}}$$ and $$x \,=\, \gamma v \tau_{\mathrm{L}}$$ into $$t' \,=\, \gamma(t \,-\, \frac{v}{c^{2}} x)$$, the expression for $$t'$$ simplifies to $$t' \,=\, \tau_{\mathrm{L}}$$. This is how you show that $$t'$$ corresponds to the time measured by the left twin's clock.
