How does an imaginary coefficient in the heat equation ensure no special direction in time?

Does the real heat equation have a special direction in time?

Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t \to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.

What does he mean by "the imaginary coefficient assures there is no special direction in time"?

The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.

The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p \to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^{ipx}$, this must be realized by $$\psi \to \psi^*$$ under time reversal, or equivalently $i \to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.

I will just paraphrase Pauli.

The Schrodinger equation has a time reversal symmetry $t \mapsto -t$ combined with complex conjugation. It looks like

$$i \partial_t \psi = \partial_{xx} \psi.$$

The heat equation on the other hand does not have a time reversal symmetry. It looks like

$$\partial_t u = \partial_{xx} u.$$

There is no way to reconcile this with the time reversed equation

$$- \partial_t u = \partial_{xx} u.$$

Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)

$$\frac{1}{\sqrt{t}} e^{-x^2/t},$$

so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like

$$\frac{1}{\sqrt{it}} e^{-x^2/it}.$$

(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t \mapsto it$.)