Using 12v 50w motorcycle light with two 7.4v RC batteries
Can you please advise the IC part number in the converter.
This will allow a much improved answer.  

IMPORTANT: If buck only, there will be a minimum Vin-Vout spec and at Vout=12V you may need Vin of > 12V.  Worst case, if (Vout-Vin)min was 4V (unlikely) the driver would be unusable for your purpose. So, the IC matters.

LiIon or LiPo cells are rated as 3.6V nominal per cell, 3V min (higher is safer) and 4.2V max.
So 2 2S 7.2V batteries in series will; give 12 - 16.8V.
Without knowing the IC used it is not possible to be sure whether it is buck only or buck boost. The single inductor , single IC suggests buck only (based on other common Chinese modules) The two pots presumably set CCCV limits.
Buck-boost with a single switch and single inductor is possible with Vin to Vout polarity inversion. 

IF correctly rate the converter has the power rating to do what you want.  

The ability to set CV &amp; CC (Vout max and I out max) makes the driver well suited to LED use.

Maximum dissipation in the driver can be expected to be ABOUT Power_in x 10-20% or about 5 - 10 Watts (maybe less) 



A less good but instant solution:

You appear to be in New Zealand.
For "playing", the use of a series/parallel resistor array will allow you to set up a driver immediately using series resistor to the LED.  This is less efficient than using the driver, dissipates much more battery energy as heat, but allows you to be experimenting within time-to-buy-from-Jaycar-NZ hours. 

A resistor voltage dropper will reduce the supplied voltage as Vbattery drops, but will compensate somewhat by increasing current somewhat as VLED drops. Overall a series resistor is an OK solution for experimenting. 

Jaycar NZ sell eg  these resistors.
4.7 Ohm 10 Watt
1 Ohm 10 watt
1 Ohm 5 Watt $0.60 / $0./42
The 5 Watt resistors cost less per Watt than the 10 Watts - 60 cents NZ each or 42 cents with a "trade card".  Buying 10 x 1 Ohm, 5W resistors allows playing (and lots of heat).

For Vin = 16.8V, Vout = 12V, Imax = 4A (48 Watts)
Use 16V. 
Rdropper = V/I = Vdrop/I = (16-12)/4 = 1 Ohms.
4 x 1 Ohm resistors in 2S2P arrangement = 1 Ohms.
More resistors in other arrangements allows you to experiment. 
All resistors dissipation = V x I =  4V x 4A = 16W = 4W/resistor.
As Vbattery drops current will fall 

Worst case resistor efficiency = Vout/Vin_max = 12/16.8~+ 75%.
A converter may achieve 80-90% and maybe better. Maybe.