Why don't these conversions of time between reference frames seem consistent?

The contradiction is in the setup. It is impossible that those three relative velocities can simultaneously be true.

You can't just add relative velocities together directly in special relativity. If an observer moving at velocity $v$ measures, in their own frame, a velocity $v'$ of some passing object, then a stationary observer would measure the velocity


If car A is moving at $0.2c$ with respect to car B, and car X is moving at $0.5c$ in the same direction with respect to car A, then car X must be moving at roughly $0.64c$ with respect to car B, not $0.7c$.

The diagram shows car X travelling at 0.7c with respects to car B and 0.5c with respects to car A. Car A is travelling at 0.2c with respects to car B. Where c is the speed of light.

This is impossible. If car X is travelling at a velocity $0.7c$ w.r.t. car B and at a velocity $0.5c$ w.r.t. car A then according to the relativistic addition of velocities, car A would be travelling at a velocity $> 0.2c$.

If the velocities of two objects are $u$ and $v$ w.r.t. an observer then the relative velocity of the first w.r.t. the second is given by


In your example, car A and car B are moving with velocities $-0.5c$ and $-0.7c$ respectively in the frame of car X. Thus, the velocity of A w.r.t. B will be given by


which would be obviously not $0.2c$ like you assume.

The morale of the story is that the Newtonian expectations of how velocities get combined don't remain valid in relativity and you need to be careful as to whether your assumptions are internally inconsistent in the purview of relativity.

Although the OP mistakenly used galilean velocity-composition ($v_{AC}=v_{AB}+v_{BC}$)
instead of special-relativisitic velocity composition ($v_{AC}=\displaystyle\frac{v_{AB}+v_{BC}}{1+v_{AB}v_{AC}}$),
using special-relativisitic velocity-composition alone
does not resolve the key issue in the OP.

The key issue is

time-dilation factors are not multiplicative

Although $\displaystyle\frac{0.5+0.2}{1+(0.5)(0.2)}=\frac{7}{11}\approx0.6363...$, $$\def\GAM#1{\displaystyle\frac{1}{\sqrt{1-{#1}^2}}} \def\BGAM#1{\displaystyle\frac{#1}{\sqrt{1-{#1}^2}}} \GAM{0.5}\GAM{0.2}\neq \GAM{\left(\frac{7}{11}\right)} $$

Instead, $\GAM{0.5}\GAM{0.2} + \BGAM{0.5}\BGAM{0.2} = \GAM{\left(\frac{7}{11}\right)}$
Here it is in a form that you can copy-paste into wolframalpha [using implied multiplication]:

Or, with an alternate set of velocities...

Although $\displaystyle\frac{0.7-0.5}{1-(0.7)(0.5)}=\frac{4}{13}\approx0.307692...$, $$\def\GAM#1{\displaystyle\frac{1}{\sqrt{1-{#1}^2}}} \def\BGAM#1{\displaystyle\frac{#1}{\sqrt{1-{#1}^2}}} \GAM{0.5}\GAM{(\frac{4}{13})}\neq \GAM{ 0.7 } $$

Instead, $\GAM{0.5}\GAM{(\frac{4}{13})} + \BGAM{0.5}\BGAM{(\frac{4}{13})} = \GAM{0.7}$
Here it is in a form that you can copy-paste into wolframalpha [using implied multiplication]:


More generally,
time-dilation factors are not multiplicative--there is an extra factor $$\gamma_{AC}=\gamma_{AB}\gamma_{BC}(1+v_{AB}v_{BC}).$$

These are easier to recognize if you work with rapidities (Minkowski angles),
where $v_{AC}=\tanh\theta_{AC}$, $\gamma_{AC}=\cosh\theta_{AC}$, $v_{AC}\gamma_{AC}=\sinh\theta_{AC}$, etc...

Since rapidities are additive $$\theta_{AC}=\theta_{AB}+\theta_{BC},$$ we have $$\begin{align*} v_{AC}=\tanh(\theta_{AC}) &=\tanh(\theta_{AB}+\theta_{BC})\\ &=\frac{\tanh\theta_{AB}+\tanh\theta_{BC}}{1+\tanh\theta_{AB}\tanh\theta_{BC}}\\ &=\frac{v_{AB}+v_{BC}}{1+v_{AB}v_{BC}}, \end{align*} $$ which is the velocity-composition formula.

So, the time-dilation-factor is $$\begin{align}\gamma_{AC}=\cosh(\theta_{AC}) &=\cosh(\theta_{AB}+\theta_{BC})\\ &=\cosh\theta_{AB}\cosh\theta_{BC}+\sinh\theta_{AB}\sinh\theta_{BC}\\ &=\cosh\theta_{AB}\cosh\theta_{BC}(1+\tanh\theta_{AB}\tanh\theta_{BC})\\ &=\gamma_{AB}\gamma_{BC}(1+v_{AB}v_{BC}), \end{align}$$ which is the formula used in the two numerical examples above.
Again, the key issue is that time-dilation factors are not multiplicative.

(You may ask
Q: "So, what factors are multiplicative?"
A: the Doppler factors: $k_{AB}=\displaystyle\sqrt{\frac{1+v_{AB}}{1-v_{AB}}}=e^{\theta_{AB}}$.)

(For a variation of this, have a look at my answer to similar question at https://physics.stackexchange.com/a/326289/148184 )