Imagine the circuit before the feedback network was installed: that is, as just a common-emitter amp, with the lower junction of emitter resistor RE and bypass capacitor CE connected to ground instead of to the feedback network.
Next consider the feedback network in isolation. Imagine looking into its input: that is, from right to left, into the two terminals on the network's right-hand side. That input will present some small impedance.
Now reinstall the feedback network: break the imaginary connection between RE//CE and ground, and insert the input terminals of the network into the gap. You've now restored the original circuit: the transistor's emitter current iE is flowing in series through the small input impedance of the feedback network. That is, the feedback network is sampling the emitter current.
That explains the "current" part of what your handout described as "current shunt feedback". (But wait: the amp's output is a voltage at its collector, not a current through its emitter. How does sampling the emitter current equate to sampling the output? A common-emitter amp delivers a voltage output by sinking collector current through load resistor Rc. The transistor's high gain means its emitter current iE is almost identical to its collector current. By sampling the emitter current therefore, effectively you're sampling the amp's output.)
Now look at the output of the feedback network - the terminal on top. A common-emitter amp embiggens a small base current to deliver a larger collector current. Before source current iS can reach the base and be amplified, the feedback network steals a small amount "if" of it - "if" is shunted away through the network's output terminal. That explains the "shunt" part of the "current shunt feedback" description.
The feedback is negative: greater emitter current - i.e. greater amplifier output - causes more of the input signal to be stolen by the network, with less remaining to reach the base and be amplified.