What is the concrete meaning of fixing an extension field through a subgroup of automorphisms in $x^3-2$?

An element $z$ is said to be a fixed point of an automorphism $\sigma$, if $\sigma(z)=z$, and that's what you are looking for here.

But I think something is wrong with the data. Let's check the action of $fr$ on the number $z=2^{1/3}(1+\omega)=-2^{1/3}\omega^2$: $$ r(z)=-r(2^{1/3})r(\omega)^2=-\omega2^{1/3}\omega^2=-2^{1/3}, $$ and therefore $$ fr(z)=f(r(z))=f(-2^{1/3})=-2^{1/3}\neq z. $$ Meaning that $z$ is not a fixed point of $f\circ r$.

On the other hand $$ r^2(z)=-r(2^{1/3})=-\omega 2^{1/3}, $$ and therefore $$ fr^2(z)=f(r^2(z))=-f(\omega2^{1/3})=-f(\omega)f(2^{1/3})=-\omega^22^{1/3}=z. $$ So $z$ is a fixed point of the automorphism $fr^2$.

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The reason Mark Bennet (in a comment) and I asked about the order of composition is that in this Galois group we have the relation $rf=fr^2$. This Galois group is not abelian, so the order of composition matters. If automorphisms were applied left-to-right, then we would have $(z)rf=z$.

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There are many ways to find fixed points for $fr$.

1. In terms of Galois theory the fixed points are related to intermediate fields. By definition $f$ fixes all the elements of $\Bbb{Q}(\root3\of2)$. By the above calculation $fr^2$ fixes all the elements of $\Bbb{Q}(z)=\Bbb{Q}(\omega^2\root3\of2)$. It stands to reason that the we should see the intermediate field generated by the third root of $x^3-2$, namely $\Bbb{Q}(\omega\root3\of2)$ also. Indeed, $$r(\omega\root3\of2)=r(\omega)r(\root3\of2)=\omega^2\root3\of2$$ and hence $$f(r\omega\root3\of2)=f(\omega)^2f(\root3\of2)=\omega^4\root3\of2=\omega\root3\of2.$$ So $\omega\root3\of2$ is a fixed point of $fr$.
2. A general fact about group actions is that if $z$ is a fixed point of $g$ then $h(z)$ is a fixed point of $hgh^{-1}$: $$(hgh^{-1})(h(z))=h(g(h^{-1}(h(z))))=h(g(z))=h(z).$$ In this Galois group we can use the relation $rf=fr^2$ I mentioned above. It implies that $$rfr^{-1}=(rf)r^{-1}=(fr^2)r^{-1}=fr.$$ Applying the general observation to $g=f$, its fixed point $z=\root3\of2$, and $h=r$, it follows that $r(z)=\omega\root3\of2$ must be a fixed point of $rfr^{-1}=fr$.
3. Then there is the lower technology, boring way, but one that is guaranteed to work. You know the effect of the automorphism $fr$ to the elements of the basis $\mathcal{B}=\{1,\root3\of2,\root3\of4,\omega,\omega\root3\of2,\omega\root3\of4\}$. You can then write the matrix $M$ of $fr$ with respect to $\mathcal{B}$. The fixed points are exactly the eigenspace of the eigenvalue $\lambda=1$. Leaving the calculations to you. That eigenspace is 3-dimensional. Given that $(fr)^2=e$, the eigenvalues satisfy $\lambda^2=1$, so $\lambda=-1$ is the other eigenvalue. The corresponding eigenspace is also 3-dimensional. That is hardly a surprise given that if $w$ belongs to eigenvalue $-1$, then so does $zw$ for all the fixed points $z$.