This comment complements the recent comment by @Matterwave ,
as well as directly connect to your recent calculation.

Mike_bb said:

Let ship A and ship B fly parallel to each other with velocity ##v=0.6c##.
Lorentz-factor: 1.25

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

This is exactly the situation in my first diagram above,
where each diamond refers to a 10-sec interval of proper-time.

##V_{Bob\ wrt\ Alice}=\frac{PP'}{OP}=\frac{6}{10}##,
where ##OP## is along Alice's worldline (maybe "Alice's timeline" is a better term),
meeting the hyperbola "Minkowski-circle" at event ##P##,
and ##PP'## is parallel to the tangent to that circle at ##P##.
Geometrically, following Minkowski (1908),
##PP'## is Minkowski-orthogonal [or "-perpendicular", "-normal"] to ##OP##.
Physically, ##PP'## trace out events that are simultaneous according to Alice.
(##PP'## could be called "a spaceline-segment according to Alice".)

Alice says the distant-event ##P'## on Bob's worldline is simultaneous with her local-event ##P##,
which is (by counting) after an elapsed time of 10 of Alice's ticks (= 100 sec).
We can count off that 8 of Bob's ticks (= 80 sec) elapses from the meeting event ##O## to event ##P'##.
So, as you say,

Mike_bb said:

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

Geometrically, note the 6-10-8 Minkowski-right-triangle (similar to a 3-5-4 one),
where ##OP## and ##PP'## are the orthogonal legs
(whose ratio ##(PP')/(OP)## gives the velocity of Bob according to Alice),
and ##OP'## is the hypotenuse (opposite the "Minkowski-right-angle at ##P##").
The time-dilation factor is ##\gamma=\cosh\theta=\frac{\rm ADJACENT}{\rm HYPOTENUSE}=\frac{OP}{OP'}=\frac{10}{8}##,
where ##\theta## is the relative-rapidity between ##OP## and ##OP'##
and ##\tanh\theta=\frac{\rm OPPOSITE}{\rm ADJACENT}=\frac{PP'}{OP}=\frac{6}{10} ## is the relative-[dimensionless-]velocity.

Bob (also being an inertial observer from event ##O##) meets the same "circle" at ##Q##,
which is also after an elapsed time of 10 of Bob's ticks (= 100 sec).
That's why this is a "circle" in Minkowski spacetime.
Bob can do a (literally) similar construction.
##QQ'## is tangent at ##Q##.
Distant-event ##Q'## on Alice's worldline is simultaneous-according-to-Bob with his local-event ##Q##.
etc... etc...
So, as you say,

Mike_bb said:

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

Time-dilation qualitatively:
With 10 ticks to get to the "circle",

• at Alice's 10th-tick, Alice's tangent line to the "circle" meets Bob's worldline
  before Bob reaches the "circle" (before Bob's clock elapses 10 ticks).
• at Bob's 10th-tick, Bob's tangent line to the "circle" meets Alice's worldline
  before Alice reaches the "circle" (before Alice's clock elapses 10 ticks).

For the Euclidean geometric analogue (the E= -1 case) , look back (above) at the ordinary circle and its tangent-lines, each meets the other radial-axis at a distance along that axis greater than the circle-radius.

In these two cases, along different radii meeting the circle, the tangents are in different directions.
This is the relativity-of-simultaneity.

For the Galilean analogue (the E=0 case), look at the desmos.
The "circle" in Galilean-spacetime is the hyperplane (in 1+1, a spatial straight-line),
whose tangent-[hyper]planes (tangent-lines) coincide.
This leads to "absolute time", with no-time-dilation.
(In Galilean spacetime geometry, the hypotenuse has the same size as the adjacent side. The underlying geometry of the PHY101 position-vs-time graph is non-euclidean (a flat non-euclidean one).)