Why is the load on a darlington pair put before the transistors rather than after?
For clarification: If you write "put the load to the high side" this means "low side switching".

In the comments section of your question there are a lot of links to the nuts and bolts of high side switching. It shows, how the complexity of high side switching looks like and where the pitfalls are. But why is it more difficult anyway? Can't you put the load simply to the emitter?

Let's compare those two circuits, assuming a forward gain of 1000 for the BJT-Darlington.
The first circuit allows up to 0.6 A on the collector in linear mode, as the voltage drop over the base resistor is 0.6V due to the two base voltages adding up to 1.4 V. This won't be reached, because the load only permits a current of 100 mA with the given voltage. 
The transistors will of course leave linear mode and enter saturation, which is not a real problem in switching application. 
The actual current through the load is only limited further by the saturation voltage which can be as low as 0.3 V with modern BJTs. So the current might rise up to 97 mA.







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The second circuit won't do what we want. Because the base current enters the node on the upper end of the load, which means for a base current to flow at all, the voltage across the load can't rise above the control voltage minus 1.4 V, which is again only 0.6 V, but this time limiting the whole emitter current to 6 mA. This is probably not, what you wanted to achieve. 







simulate this circuit