Subtracting standard electrode potentials As far as I'm aware, if we construct a half cell ($\ce{A + e- &lt;=&gt; A-}$) under standard conditions then the standard electrode potential $E^⦵$ is the potential of the $A/A^{+}$ couple wrt. SHE, which doesn't change depending on whether the reaction is an oxidation or reduction in this half cell. This is because potential is a physical, electrostatic concept. 

Consequently, the cell EMF is then also defined to be the potential of the cathode subtract that of the anode, $E^⦵_{cell} = E^⦵_{cat} - E^⦵_{an}$. All of the above I have obtained from IUPAC recommendations.

However, some people say that reversing a half equation reverses the sign of $E^⦵$, and then proceed to add the electrode potentials to find the cell EMF. Although this method is algebraically equivalent, it doesn't make any sense - since the electrode potential has nothing to do with the chemical concept of a reaction direction!

The only justification I have seen is that if you write $\Delta G = -nFE^⦵$ for each half equation, and sum them, then the EMF comes out as expected only if we do change the sign. However this seems inherently flawed as well, since although Gibbs free energies are additive, $\Delta G = -nFE^⦵$ only applies to a cell as a whole (because it is derived by considering the energy change if charge flows from one terminal to the other) and is meaningless in the context of a half equation.

Consequently, why is the notion that reversing the half cell equation negates the electrode potential still taught - since to me at least it seems fundamentally incorrect!

Thanks a bunch!
Let's take a look at what you're doing specifically for computing the potential difference. The potentials at the cathode and the anode are all relative to some standard. Importantly, it must be the same standard. Let's refer to this as ground.

$$E_{\mathrm{cathode}} = V_{\mathrm{cathode}} - V_{\mathrm{ground}}$$
$$E_{\mathrm{anode}} = V_{\mathrm{anode}} - V_{\mathrm{ground}}$$

The $V$'s here represent absolute potentials on some arbitrary scale.

What's the change in potential when I move a test charge from the anode to the cathode? Since potential differences are path independent, we can look at this via states: moving the charge from anode to cathode is the same as moving the charge from anode to ground and then from ground to cathode.

The potential difference is expressed as:

$$E = (V_{\mathrm{cathode}} - V_{\mathrm{ground}}) - (V_{\mathrm{anode}} - V_{\mathrm{ground}}) = E_{\mathrm{cathode}} - E_{\mathrm{anode}}$$

The key is the cancellation of the ground potential, so that we have a potential difference.

This is exactly what are doing when working with half-reactions. And it works because both the reduction and oxidation half reactions are referenced to the standard hydrogen electrode.

Let's look at two half-reactions:
$$\ce{A -&gt; A+ + e-}\ ,\ \ E^{⦵} = -E_{A}$$
$$\ce{B+ + e- -&gt; B}\ ,\ \ E^{⦵} = E_{B}$$

As a general rule, we tabulate reduction potentials, so let's treat the $\ce{A}$ reaction with a reduction potential:

$$\ce{A+ + e- -&gt; A}\ ,\ \ E^{⦵} = E_{A}$$

Now, both of these half reactions are relative to the standard hydrogen electrode.

In the reaction
$$\ce{A + B+ -&gt; A+ + B}$$
we can look at the equivalent process of moving an electron from $\ce{A}$ to a proton in a standard hydrogen electrode, and then taking a electron from dihydrogen in the standard hydrogen electrode and moving it to $\ce{B}$.

The potential difference is:
$$E^{⦵}_{\mathrm{cell}} = E_{B} + (-E_{A}) = E_{B}-E_{A}$$

This is exactly the same process because the anode potential relative to ground is the reduction potential of the oxidation half-reaction. The potential difference is computed by subtracting the potential from the other one.

Alternatively, if you already expressed the anode potential as a oxidation potential (inverse of the reduction potential), then the minus sign is implicitly included, and you can just add the two potentials.