what determines the lower limit on the number of turns in a secondary coil in a step down transformer?
I am interested in understanding what factors govern the lower limit on the number of turns in the Secondary coil of a step down transformer. 


Good. 


  I have read that a Primary coil with too few turns will induce a very weak magnetic field in the core,


You have remembered wrongly, or misread. 

You need at least a minimum number of turns on the primary in order to keep the core field below its maximum.

Core materials, iron or ferrite, have a maximum magnetic field. With iron, the limit tends to be saturation, with ferrite, heating from losses tends to bite before saturation. The maximum fields are typically in the order of 1.5T for mains iron transformers, and in the range 0.1T to 0.3T for ferrite depending on frequency and grade.

The core has an area \$A\$, and the core material has a maximum field \$B_{max}\$. The transformer has an operating frequency \$f\$. We can calculate the maximum voltage induced in one turn via the maximum rate of change of flux, which is \$V_{max} = 2\pi f B_{max} A\$ for a sinusoidal input, with all terms in SI units. This gives us the peak voltage. Divide by sqrt(2) to give the RMS voltage. We need enough turns to equal our input voltage. 

To give an example, I have an iron toroid with cross section 13x25mm, operating at 50Hz, and I am going to assume a \$B_{max}\$ of 1.5T.

\$V_{max} = 2 \pi \times 50 \times 0.013 \times 0.025 \times1.5 = 0.153V  \$

0.153V peak is about 0.108V rms. At 240v, we would need 2222 turns on the primary, minimum. A few more wouldn't hurt.


  but what happens if the secondary coil has a very low number of turns?


then you get a very low secondary voltage produced.


  If the secondary coil uses a shorter wire, the resistance is lower and so voltage induced in the secondary coil produces a greater current, as per Ohm’s law. 


well, yes, if you shorted the secondary. But we don't do that. The resistance of the load tends to exceed that of the winding, and so it's the load that determines the current.

You do several things to design a transformer. Note that a full design could get much more detailed, but this gives you a good start.

1) Start with a big enough core for your power. Go to a transformer catalogue, and find a transformer with a similar power to what you want to use. Halve its weight, the copper is about the same weight as the iron. Start with a core this weight.

2) Calculate the minimum number of turns needed on the primary, as above

3) Caluclate the number of turns needed on the secondary, to give your design voltage, using the same volts per turn figure.

4) Choose a the maximum thickness of wire that will half-fill the winding window. 50% is a good fill factor, you will be doing well to get a better factor.

5) Check that the current density doesn't exceed about 3A/mm2 (a rule of thumb valid for small to medium transformers) at your load power.