This isn't the time dilation aka rate change, but rather due to perspective change, where is Sam looking on Sally's worldline?
For example, imagine Sam is 1000 lightyears from Sally and Sally is receding at 0.1 the speed of light when Sam instantly accelerates to 0.5 the speed of light toward Sally. What equation describes what Sam observes regarding the time jump reflected on Sally's watch?
Also, are there any good physics GitHub repos that do a good job of modeling this stuff?
Addressing the Time Jump:
The equation for relativity of simultaneity is: $$ t' = \frac{t-v x}{\sqrt{1-v^2/c^2}}.$$
Here $t'$ is the time seen on a clock at distance $x$ that is moving at velocity $v$ relative to the observer.
In Sam's case when he arrives at the turnaround, $6 \ \text{years}$ have passed on his clock, so $t=6$. The turnaround is $8 \ \text{lightyears}$ from Earth but due to length contraction Sam considers the distance to be $$x = \frac{8 \times 6}{10} = 4.8 \ \text{lightyears}.$$
Using the simultaneity equation Sam calculates the time of Sally's Earth clock to be $$\frac{6-0.8 \times 4.8}{\sqrt{1-0.8^2}} = 3.6 \ \text{years}.$$
After the turn around he calculates the time on Sally's clock to instantly read $$ \frac{6+0.8 \times 4.8}{\sqrt{1-0.8^2}} = 16.4 \ \text{years}.$$ This is the time 'jump'. Sally's clock has instantly jumped forward by $12.8 \ \text{years}$! This is not what he actually sees and is not physical. It is an artifact of Sam suddenly switching reference frame and it is normal for observers in different reference frames to measure things differently.
On his return trip he sees Sally's clock advance a further $3.6 \ \text{years}$ making a total of $20 \ \text{years}$ on Sally's clock as expected.
Why the time jump occurs.
Imagine Sam has very long spaceship. In fact it is so long, that even when the front of the ship arrives at the turnaround point $8 \ \text{lightyears}$ away, the back of the spaceship is just passing the Earth. Now initially Sam had clocks synchronised in his moving rocket, but to Sally on Earth, the clock at the back appears to be much further advanced in time than the clock at the front of Sam's ship. Now Sam instantly reverses the the ship to return to Earth.
Here is the key point: clocks do not re-synchronise themselves when a reference frame changes velocity. Sam has to have someone at the back adjust the rear clock to make it synchronised to the front clock. Let's say it is Betty (at the back)'s job to do this. She retards her clock manually by $12.8 \ \text{years}$ to re-synchronise with Sam's clock. Betty is right next to Sally at this point, and Sally sees Betty reverse the clock by $12.8 \ \text{years}$, so Sally is not surprised that there is sudden time jump because it was done manually. Betty is not surprised that Sally's clock has instantly advanced by $12.8 \ \text{years}$ because she just adjusted the clock. Nothing mysterious happened. The jump is purely artificial bookkeeping artifact.
The equation for the relativistic Doppler effect is: $$ f_r = f_s \sqrt{\frac{1-v_s/c}{1+v_s/c}}$$ where $f_s$ is the frequency of the source and $v_s$ and $f_r$ are the velocity of the source and the frequency as measured by the receiver. The observed frequency is basically the observed tick rate of a distant clock.
In Sam's case he sees Sally as going away at 0.1c so he measures her clock to slower than his clock by a factor of $ f_r = f_s \sqrt{\frac{1-0.1}{1+0.1}} \approx 0.9045340 \ f_s $.
Sally also measures Sam to be moving away at 0.1c and so she also measures Sam's clock to be ticking slower by the same factor.
What equation describes what Sam observes regarding the time jump reflected on Sally's watch?
There is no "jump" as such. If Sam sees 3:30 PM on Sally's watch and instantly accelerates to 0.5c, he will still see 3:30 PM on Sally's watch immediately after the boost. What does change is the rate at which he sees Sally's clock tick at before and after the boost. Now Sam effectively sees Sally velocity as 0.1 - 0.5 = -0.4c, i.e. he Sally coming towards him now. The tick rate of Sally's watch that Sam sees after the boost is $ f_r = f_s \sqrt{\frac{1+0.4}{1-0.4}} \approx 1.52753 \ f_s $. To him, Sally's watch instantly appears to be ticking significantly faster than his own watch after the boost. Sally on the other hand will not see a change in Sam's tick rate for another 1000 years due to the light travel travel distance, but eventually she will also see Sam's watch tick faster by the same factor.
Because they are both travelling towards each other they will eventually pass each other and after they pass they will once again see each other watches ticking slower than their own watches, but this time by a factor of $ f_r = f_s \sqrt{\frac{1-0.4}{1+0.4}} \approx 0.654654 f_s$.
Twins Paradox Calculation:
In the comments the OP asked how this applies to the twins paradox. Here is a worked example:
Imagine Sam travels way from Earth at 0.8c for 10 years of Earth time. In that time he travels 8 light years. He sends signals every year according to his clock. The signal he sends on arriving at the turnaround does not get to Sally until 8 years after he arrives at the turnaround. Sally sees this turnaround signal after 18 years Earth time because it takes Sam 10 years to get to the turnaround and the signal takes 8 years to return. Due to the Doppler shift, Sally sees $ 18 * \sqrt{\frac{1-0.8}{1+0.8}} = 18/3 = 6 $ birthday signals from Sam on his outward journey.
For the last 2 years after the turnaround Sally sees $ 2 * \sqrt{\frac{1+0.8}{1-0.8}} = 2*3 = 6$ birthday signals so she sees Sam age by a total of 12 years during her 20 years waiting on Earth.
On arrival at the turnaround Sam has aged by 6 years by his clock. In that time he sees $ 6 * \sqrt{\frac{1-0.8}{1+0.8}} = 6/3 = 2 $ birthday signals from Sally. On the return trip he sees $ 6 * \sqrt{\frac{1+0.8}{1-0.8}} = 6 * 3 = 18 $ birthday signals from Sally. Altogether he sees 20 signals from Sally which agrees with the time Sally actually spent waiting on Earth.
The relativistic gamma factor at 0.8c is $1/\sqrt{1-0.8^2} = 10/8 \approx 1.6666$.
20 years/1.66 = 12 years in agreement with the earlier calculations.
