Circuit wanted to run 12VDC LEDs from Low Voltage landscape circuit
In order of increasing preference:


1 A series diode to a capacitor 
2  A bridge rectifier to a capacitor. 
3 A bridge rectifier to capacitor, with a series resistor to LEDs (best, a series resistor in each string). 
4 Overkill: Bridge rectifier, capacitor, series regulator or resistor + shunt zener


LED strings in parallel in each case.

Because: The "12 Volt" weatherproof LEDs are almost certainly between 1 and 3 LEDs in series plus  a resistor. This combination allows some tolerance as increasing applied voltage causes increasing current which leads to more drop across the series resistor, so the LEDs voltage drop increases only "modestly" with increasing voltage. 

Option 3 is liable to be the most effective.
If the LDs are standard single 5mm or 3mm LEDS the current drain is probably designed at 20 mA. If they are series parallel strings then they could have almost any rating and you need either to look at a data sheet, or measure current at 12V or use trial and error.

The 12 VAC supply, when full wave rectified with a "bridge" rectifier, will produce about 16 VDC. Actual voltage depends on length of wiring, load from other lights, mains voltage fluctuations etc BUT 16 VDC is a good first guess. So you need to drop about 16-12  = 4 volts. Series resistor required is \$ R = \frac{V}{I} \$. At 20 mA this would be \$ R = \frac{4}{0.02} \approx 200\Omega\$. Nearest standard "E12" values are 180R (180 Ohms) or 220R (220 Ohms). Chances are that 180R would be fine and even 270R or 330R or 150R or 120R would be good enough. 

The capacitor needs to be large enough to reduce ripple from raw rectified AC and as small as possible for $ and size reasons. The whole arrangement can be placed in a minibox or potted in epoxy or ... . For outdoor protection I sometimes use a pair of suitably sized soft-drink bottles cut such that one pushes over the other and pressed together and located so water has a "hard job" entering. This can be as capable and aesthetic an arrangement as your skill and caring allow.

The bridge rectifier can be almost any such - small ones rated at 100's of mA to 1A to much more are available from  electronic components suppliers (Frys?, Radio Shack?, Digikey, Farnell, Mouser, ...).

Power dissipation is mainly in the resistor and is \$I^2 * R\$ or \$\frac{V^2}{R}\$  or \$ V * I \$.
(V = voltage drop across R, I = LED current.)
e.g. for 180R and 20 mA Power:  


  \$ R = I^2 * R = 0.02^2 * 180 = 0.072 \;\;\$  Watts = trivial.  


If your 12V LED is e.g. 3 x 1 Watt rated at 350 mA then power in series:  


  \$ R = V * I = 4 * 0.35 = 1.4 \;\;\$ Watts.  


This is less trivial. Potting that in epoxy without care may led to "problems". A suitably power rated resistor is required in each case.

Capacitor is sized to allow acceptable ripple during about one half mains cycle. Roughly:


  Voltage Drop \$ = V  = t * \frac{I}{C} \;\;\$ Volts
   (t = seconds, V = drop in time t, I = current drawn, C = capacitance  in farads. 


Rearranging \$C= t * \frac{I}{V}\$
For 60 Hz t ~= 8 mS, i = 20 mA say, V = 1 Volt drop say.
C = 0.00016 F = 160 uF.
i.e. you need ABOUT 160 uF per 20 mA LED. 
You MAY be happy with no cap at all and full wave DC on LEDS. You MAY see too much flicker with e.g. 100 uF per 20 mA LED.
Chances are that a 1000 uF, 16 Volt capacitor will be fine for 3 x 20 MA LEDS and that smaller will be OK.

Sensibility check After building, measure Vdc across LEDs . Should be ~~~~ 12 Volts.

All the above said, solution 1. or 2. may work well enough for you.

Added: The extra information and page you referred to suggests that they may use a SR (switching regulator) (probably a buck converter) internally. The ones shown state that they are rated for 8 to 24 V, which is a very wide range for a linear regulator due to efficiency considerations and far beyond what you can sensibly achieve with a resistor only. The current rating of 156.3 mA at 12V to give 71 lumen output also strongly suggests a SR.
12v x 156 mA = 1.9 Watt or 38 lumen/Watt. This is modest but realistic in real world applications. 

So 


Contact the suppliers and ask them and/or
Try the 2nd circuit with no cap. Flicker may be objectionable.
Add a capacitor as in 2 &amp; 3. At 156 mA, using my 160 uF per 20 mA rule of thumb, for 1 Volt drop you'd want 


156 mA / 20 mA x 160 uF = 1248 uF, so a 16V, 1000uF aluminium electrolytic would probably be fine.