How is a transconductance amplifier used to measure current?
This is not so much an answer, since answers have already been given, as an addition that is to long for a comment.


  a transconductance amplifier (again with programmable gain) which
  produces a voltage proportional to the current through it


The author is confused or sloppy.  A transconductance amplifier converts a voltage input to a current output:

$$i_{out} = g_mv_{in}$$

As the other answers imply, a transresistance (transimpedance) amplifier converts a current input to a voltage output:

$$v_{out} = r_mi_{in}$$

As you may know, an ideal transresistance amplifier has zero input resistance (and zero output resistance) so the input current does no work on (delivers no power to) the input.  I suppose this is what is meant by 'burden'.

The ideal (ideal op-amp etc.) circuit George Herold gives has zero input resistance and zero output resistance.