Light will travel 10 light-seconds (2,997,924,580 meters) in 10 rail-seconds for an observer on the rails according to his own ruler.
Light will travel 10 light-seconds (2,997,924,580 meters) in 10 train-seconds for an observer on the train according to his own ruler.*
Yes, the light will travel 3 000 000 kilometers in 10 rail-seconds for an observer on the rails. The coordinates of the light (x; T) are (3 000 000 km ; 10 rail-second) 
According to Einstein, the tempo of relative system's time is increased because of relative speed of material or system. The unit of time has dilation and the time deals slower. We have known system of the measurement for time as second, minute, hour, day…..
If the interval of known/reference second as unit is "one", the value of the relative second as unit is 1/[1 - (v/c)^2]^1/2. This value is the ratio of known/reference second.
If v = 60 % c ; 1 unit of relative second = 1.25 reference second
A photon which it is chosen as a proceeding actor is a singular reality or existing. In other words it has unique position at any moment. The time of this position may be defined by different units of time. But the results appear this single existing. In other words the coordinates of light x, y, z, T and x', y', z', T' mark the unique point in space. The moments of T and T' are the codes of the same moment by different units of time.
So, The observer in the train will see on his chronometer
T' =Total time / his relative unit of time = 10 rail second / 1.25 rail second = 8
The observer must read 8 train-second on his choronometer. And The same light travels 2 400 000 train-kilometer for this time
[ If the light travels 10 seconds, how many relative seconds does the same photon travels in relative system?
t' = Total time according to reference unit of time/ the value of unit of relative time by reference unit.
In our numerical example (v = 0.60.c): t' = 10 / 1.25 = 8 relative seconds.
The monitor of relative chronometer is read 8 second. It means 8 relative seconds ]
The relative coordinates (x' ; T') are (2 400 000 Km ; 8 train-seconds) according to Einstein.
But Lorentz transformings give above values for relative coordinates (x' ; T'):
x' = (x - v.t) / [1 - (v/c)^2]^1/2 = 1 500 000 train-km
T' = (t - v.x/c^2) / [1 - (v/c)^2]^1/2 = 5 train-seconds 
You said that:"You don't understand the SR". It means You, dear Modest claims to understand better. OK. Please explaine the different results for the coordinates  and .*
Note for visitors: YES this answers are correct, if two different experiments are organized by two independent actor of light.. In fact they seem right by careless mode. But we must not allow the superficial sayings lead our logic. In other words we may/must organize the sentences perfectly. NO
, these sayings have not precision/perfection. The "measuring" is precision; but "traveling" is not certain. To choose the code "traveling" is our logic's defect; it is not absolutely; it requires examination. That statement may be better: If we measure the velocity of light on the rails or in the train we will get always the velocity according to most external reference system, because our measuring system does never measure the relative velocity according to local frame**
. ** We can pass over some technical difficulties to measure the relative velocity of light according to local frame: For example, we may use to determinate the moments of time the technology of camera instead of the system by perforated wheel. In deed we must the use the actor of light as flashing point or a single photon and without mirrors. We have a high performance camera (detector) for the moment Tı. The first mark on the film gives the moment Tı. The moment of flashing To is easier. t = Tı - To. We will calculate L = c.t . If the distance between flashing point and the surface of film is L SR is right absolutely. But in my opinion probably the distance L < c.t or >c.t and L' = (c +/-V) t ; L' is the length of road of the traveling by light de facto.