You can place the RC either at the B side or the A side. When components are placed in series the order of them doesn't matter for the working.
About the diodes. When you switch off the relay it will cause a (possibly large) negative voltage on the FET's drain, and a flyback diode is used to limit that voltage to a 0.7 V diode drop. So the diode(s) don't serve to protect the coil, but the FET. Using the zeners will allow this voltage to go to -5.7 V or -15.7 V if you'd use the 15 V zeners. There's no reason for taking risks here, even if the FET can handle -30 V. So I would just use a rectifier or signal diode, or even better a Schottky diode.
edit re your comment
You can indeed use a zener (combined with a common diode, D1 doesn't have to be a zener) to decrease switch-off time, and Tyco also mentions it in this application note, but I don't read it as if they insist on it. The scope images in the first link show a dramatic decrease in switch-off time, but that measures the time between deactivating the relay and the first opening of the contact, not the time between first opening and the return to the rest position, which will change much less.
edit re the 6 V relay and the RC circuit
Like I says in this answer you can operate a relay below its rated voltage, and since its operate voltage is 4.2 V the 6 V version of your relay can also be used at 5 V. If you use a series resistor not higher than 9 Ω you'll have that 4.2 V, and then you don't need the capacitor (keep an eye on the tolerance for the 5 V!). If you want to go lower you're on your own; the datasheet doesn't give a must hold voltage. But let's say this would be 3 V. Then you can use a series resistor of 32 Ω and you'll need the capacitor to get the relay activated.
Operate time is maximum 15 ms (which is long), so as the capacitor charges the relay voltage shouldn't go below 4.2 V until 15 ms after switching on.
Now we have to calculate the RC time for that. R is the parallel of the relay's coil resistance and the series resistance (that's Thévenin's fault), so that's 19.3 Ω. Then
\$ 3 V + 2 V \cdot e^{\dfrac{- 0.015 ms}{19.3 \Omega \text{ C}}} = 4.2 V \$
Solving for \$\text{C}\$ gives us 1500 µF minimum.
Re switching off:
You can't violate Q = CV, it's the Law. Your clamping voltage is 3.3 V + 0.7 V = 4 V. That means that when you switch the FET off the low side of the capacitor momentarily will be pulled to -4 V, and quickly rise again to 0 V. The high side is 2 V higher, and will simply follow that 4 V drop while the capacitor discharges through the parallel resistor. The capacitor won't even notice the drop. The discharge time constant is 1500 µF \$\times\$ 32 Ω = 48 ms, then the capacitor will discharge to 20 mV (1% of its initial value) in 220 ms.
The 62 mA won't charge nor discharge the capacitor. We often apply Kirchhoff's Current Law
(KCL) to nodes, but it also applies to regions:
Draw a boundary around C1 and R1, and you'll see there's only one path to the outer world since the way to the FET is cut off. Since the total current has to be zero there can't be any current through that unique connection. The coil has to take care of the 62 mA on its own, and it does so by using the loop formed by the zeners.
It is because the fan is an inductive load. The relay is rated for 3A resistive load. Sometimes you will be switching the relay off when the current through the fan motor is at a peak, and hence you will get a high voltage surge - causing the contacts to arc, this tends to weld the contacts together.
I suggest you use a solid state relay with zero crossing switching for this instead.
Best Answer
simulate this circuit – Schematic created using CircuitLab
It reads as though you have created Circuit 1 above. The relay coil will limit the current through the lamp. I think you intended Circuit 2.
Circuit 2 is not much good as it doesn't prove that the light is connected or working.
Please post some details (in your original post) on the LED navigation lamps (current, voltage, wattage) with links to data sheets and we can suggest solutions.
Update after details added to question
Executive summary
Replace your 24 V relay with a 12 V unit. You'll need to add some parallel resistance to pass enough current to light the lamp brightly.
Calculations
We know that LEDs brightness is controlled by current. The fact that the light operates from 9 to 33 V suggests that they've got a current regulator in there. We can figure out what current it's drawing by using the 2.5 W power and the 9 V specification.
Since $$P = V·I$$ we can calculate the current required as $$I = \frac{P}{V} = \frac{2.5}{9} = 0.28 A$$
Your relay-in-series with lamp approach is the most reliable as if the wires fall off the lamp or it fails open circuit the relay will drop out so let's see if we can make it work.
If we share the 24 V giving half to the lamp and half to the relay we will have the following.
simulate this circuit
We need 0.28 A going through the LED so we need a relay and maybe R3 to pass 0.28 A as well. From Ohm's law we can calculate that the resistance of the coil and R3 in parallel should be $$R = \frac{V}{I} = \frac{12}{0.28} = 42 Ω$$
Since most suitable 12 V relay coils would have a resistance higher than this we will need R3 in parallel.
$$R3 = \frac{42·Rc}{Rc - 42}$$
Example: 12 V, 400 Ω relay would require $$R3 = \frac{42·400}{400 - 42} = 46 Ω$$
Finally we need to work out the power rating of R3.
$$P = V·I = 12·0.28 = 3.36 W$$
The nearest stock sizes in this example would be 47 Ω, 4 W wirewound. Note that the resistor will get hot.