I will use the more traditional notation $\frac{\mathrm D}{\mathrm Dt}$.
$$\frac{\mathrm D}{\mathrm Dt}(\boldsymbol \omega\cdot\nabla f)=\boldsymbol \omega\cdot\frac{\mathrm D}{\mathrm Dt}(\nabla f)+(\nabla f)\cdot \frac{\mathrm D\boldsymbol \omega}{\mathrm Dt}$$
Let's break this down:
$$\frac{\mathrm D}{\mathrm Dt}(\nabla f)=\frac{\partial}{\partial t}(\nabla f)+(\boldsymbol u\cdot \nabla)(\nabla f) \\ =\nabla(\partial_t f)+(\boldsymbol u\cdot \nabla)(\nabla f)$$
Breaking into components using index notation,
$$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_j\nabla_k f$$
We live in flat space, so we can commute the covariant derivatives, and so
$$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_k\nabla_j f$$
For a moment consider
$$\nabla_k(u^j\nabla_j f)=u^j\nabla_k\nabla_j f+(\nabla_j f)(\nabla_k u^j)$$
Therefore,
$$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_k\nabla_j f=\nabla_k(\partial_t f+u^j\nabla_j f)-(\nabla_k u^j)(\nabla_j f)$$
In other words,
$$\frac{\mathrm D(\nabla f)}{\mathrm Dt}=\nabla\left(\frac{\mathrm Df}{\mathrm Dt}\right)-(\nabla\boldsymbol u)\cdot(\nabla f)$$
So the relation
$$\frac{\mathrm D(\nabla f)}{\mathrm Dt}=\nabla\left(\frac{\mathrm Df}{\mathrm Dt}\right)$$
Holds iff $\nabla f=0~,~\nabla\boldsymbol u=0$, or both.
EDIT:
Holds iff $\nabla f=0~,~\nabla\boldsymbol u=0$, or both.
Wow, this is dumb. Don't know why I said that. It is (obviously) possible for the dot product to be always zero even when neither of the fields are always zero.
Application: If we have an incompressible linear isotropic fluid with no external forces, it obeys the INSE:
$$\frac{\mathrm D\boldsymbol u}{\mathrm Dt}-\nu\nabla^2 \boldsymbol u=-\nabla p \tag{1A}$$
$$ \nabla\cdot\boldsymbol u=0\tag{1B}$$
If the fluid happens to be irrotational as well, then there exists a scalar field $\phi$ such that $\boldsymbol u=\nabla\phi$. Plugging this into $(1\mathrm B)$, we get
$$\nabla\cdot\nabla\phi=\Delta\phi=0$$
So $\phi$ satisfies Laplace's equation. Additionally, plugging this into $(1\mathrm A)$,
$$\frac{\mathrm D(\nabla\phi)}{\mathrm Dt}-\nu \Delta (\nabla\phi)=-\nabla p$$
We already have an expression for $\frac{\mathrm D(\nabla\phi)}{\mathrm Dt}$, and since in flat space, we have $\nabla\Delta\phi=\Delta\nabla\phi$, we get
$$\nabla\left(\frac{\mathrm D\phi}{\mathrm Dt}\right)-(\nabla \phi)\cdot(\nabla\boldsymbol u)-\underbrace{\nu\nabla(\Delta \phi)}_{=0}=-\nabla p$$
We switch to index notation now:
$$\nabla_i\big(\partial_t \phi+u^j\nabla_j\phi\big)-(\nabla_j\phi)(\nabla^j u_i)=-\nabla_i p$$
Substitute $u_k=\nabla_k\phi$:
$$\nabla_i\big(\partial_t\phi +(\nabla^j\phi)(\nabla_j\phi)\big)-(\nabla_j\phi)(\nabla^j\nabla_i\phi)=-\nabla_i p$$
And again, in flat space, we can switch the order of the derivatives, i.e $\nabla^j\nabla_i=\nabla_i\nabla^j$, so the above becomes
$$\nabla_i\big(\partial_t\phi +(\nabla^j\phi)(\nabla_j\phi)\big)-(\nabla_j\phi)~\nabla_i\big((\nabla^j\phi)\big)=-\nabla_i p$$
Expanding the left derivative,
$$\nabla_i\big(\partial_t\phi+(\nabla^j\phi)(\nabla_j\phi)\big)=\nabla_i(\partial_t\phi)+\nabla_i\big((\nabla^j\phi)(\nabla_j\phi)\big) \\ =\nabla_i(\partial_t\phi)+(\nabla^j\phi)\nabla_i(\nabla_j\phi)+(\nabla_j\phi)\nabla_i(\nabla^j\phi)$$
Hence our momentum equation is now
$$\nabla_i(\partial_t\phi) +(\nabla^j\phi)\nabla_i(\nabla_j\phi)=-\nabla_i p$$
And since
$$(\nabla^j\phi)\nabla_i(\nabla_j\phi)=\frac{1}{2}\nabla_i\big(|\nabla\phi|^2\big)$$
We get
$$\nabla\left(\partial_t\phi+\frac{1}{2}|\nabla\phi|^2\right)=-\nabla p$$
And finally, under the assumption that the fluid does no work, this becomes Bernoulli's equation(s) for incompressible fluids
$$\partial_t\phi+\frac{1}{2}|\nabla\phi|^2+p=0\tag{2A}$$
$$\Delta\phi=0\tag{2B}$$
