In an inertial frame K, duration can be measured at each single location by using a uniformly ticking clock that is fixed right there.
Note: uniformity is ensured by producing all ticks by the very same method, so that we cannot think of any reason why one tick should take longer than the other.
The simultaneity of events that happen directly to entities at rest relative to K is measured as follows:
(Definition) Two events are simultaneous in K if and only if a symmetrically placed observer in K sees them, by the naked eye (through vacuum), happen simultaneously.
This definition is compatible with the classical conception of time.
Notes: (a) by saying "observer in K", it is meant that the observer is also at rest relative to K, (b) we can imagine that each entity sends a light signal to the observer when its local event happens, (c) here and in the following, locations, events, entities, observers, signals and clocks are, unless explicitly stated otherwise, always meant to be point-like, (d) accordingly, by "light signal" it is understood just the tip of a ray of light.
Instead of light, other signals could also be used, e.g. pistol bullets or even carrier-pigeons, as long as the symmetry of their propagation can be assumed. Since we are talking about mechanics, it seems reasonable to require only the symmetry of the net forces acting upon the chosen pair of (identical) "messengers". Einstein's original approach eliminates this requirement elegantly: it uses light signals and assumes that nothing whatsoever can make a difference to their propagation in vacuum. The same is definitely not true for e.g. a bullet whose motion is affected by a multitude of gravitational forces at the very least.
(Definition) Two clocks (at rest) in K are synchronized if the same positions of their hands are simultaneous events in K.
Notes: (a) in general, by saying "E in K" it is meant that E is at rest relative to K, (b) we'll always assume that all clocks are of the same construction.
Using mostly symmetry-based reasoning, we infer all the below:
(Theorem) If one position of the hands of two clocks in K are simultaneous events, then all positions are.
(Proof) There is no such difference in the situations of the two clocks that would explain why a symmetrically placed observer would see them showing different times, ever.
(Lemma) If a light signal is sent earlier from one location to another in K, it also arrives earlier.
(Proof) If synchronized clocks are used at the two locations, the difference between the time values at sending and arrival, as shown by the corresponding co-located clocks, is always the same. This can be seen if we add another pair of synchronized clocks that, when sending the second light signal, both show the time as it was when the first one was sent. The original and the added clocks run then in parallel, and there is no reason why the latter would not show the same time difference for the second light signal as it was for the first one.
(Theorem) All symmetrically placed observers in K judge the simultaneity of two events the same way.
(Proof) Let o1 and o2 be two symmetrically placed observers, and let o2 send light signals s1 and s2 to o1 upon seeing the two events e1 and e2, respectively. Due to symmetry, o1 will observe the same time difference between seeing s1 and e1 as that between seeing s2 and e2. Thus e1 and e2 are simultaneous to o1 if and only if s1 and s2 are sent by o2 at the same moment.
(Theorem) Simultaneity in K is transitive.
(Proof) Let event e1 be simultaneous with e2, and e2 with e3. If any two of the events are co-located, the statement is trivial. Otherwise, create an event e4 which is simultaneous with e2 but not located on any of the e1e2, e1e3, e2e3 lines. Then, an observer in the circumcenter of the triangle e1e2e4 will see that e1 and e4 are simultaneous. Similar is true for the triangle e4e2e3, i.e. e4 and e3 are simultaneous too. Finally, an observer in the circumcenter of the triangle e1e4e3 will see that e1 and e3 are simultaneous.
The last two theorems basically say that the definition of simultaneity is consistent.
For simplicity, it is assumed in the following that in every inertial frame, all clocks are already ticking in sync… just by coincidence.