Here is a possible non-microcontroller solution.
SW1 and SW2 are the two reed switches; let's say SW1 is the "speed reed" and SW2 is the "angle reed". When a reed switch closes, after a short time (determined by RSn and CDn, the debouncing components) the input of the relevant Schmitt trigger inverter gate will be regarded as low, and the gate's output will go high.
This high-going edge will be passed through CTn and will produce a short positive pulse across RTn, which will pass through DTn and will force the U1C/U1D latch into the appropriate state.
Therefore, the latch is set on active-going transitions of one reed switch, and the latch is reset on active-going transitions of the other reed switch.
The output of the latch therefore represents the time between activations on the reed switches.
This voltage is smoothed by RS and CS, which have a time constant of about 1 second, and the resulting voltage is the output of the circuit, labelled WD (wind direction).
The WD output represents the wind direction. Due North corresponds to 0V and 5V (see below) and due South corresponds to 2.5V. East and West correspond to 1.25V and 3.75V; which way around depends on details of the design of the anemometer which are not specified in the document you linked to. You will have to find this out through trial and error.
As I've mentioned in earlier posts, there are a few problems with this approach.
Ideally, the RS/CS smoothing circuit should adapt its time constant to suit the actual rate of the pulses from the reed switch; this is not simple to do, so I haven't tried to show it here. Provided that the wind speed is high enough to produce at least ten edges on each reed switch per second, the ripple on the WD signal should be reasonably low.
If the wind vane is pointing roughly North and moving slightly, the phase relationship between the two reed switches can jump from one leading the other, to the other leading the first. This will cause the WS output to alternate between 0V and 5V as the lead/lag relationship changes; this would seem to indicate that the wind direction is changing rapidly, but that's not the case.
The output smoothing circuit is a simple resistor-capacitor averager. These take a long time to settle fully to a new value. They move towards the new averaged value at a rate that depends on the difference between the ideal averaged value and the current value; as the averaged value approaches the ideal value, this difference becomes small, and therefore the rate of change also becomes small.
The voltages I've stated assume that U1, the CD40106B, is running from a 5V supply. It will actually run better from a higher voltage; 12V is good. In this case, output voltages will be scaled up accordingly.
U1 can also be a CD4584 or a 74C14. These are just different names for exactly the same IC. You can also use a CD4093B if you tie the unused inputs high.
If you build this circuit, please let me know how it goes. I suspect it may not be good enough for your requirements, but you haven't really stated those, so I can't be sure. As I said before, I think the best solution to this problem is a microcontroller.
Edit:
1. The capacitors marked "FILM" should be metal film or similar types, not ceramic - especially not MLCC (multi-layer ceramic).
2. CS should be shown as an electrolytic, with its top plate being positive.
3. The power connections to U1 are not shown. VCC connects to pin 14 and 0V (the earth symbol) connects to pin 7. Connect a 100 nF decoupling capacitor across those pins; in this case a ceramic or MLCC is fine.