In the following, I will have a conclusion that is definitely wrong but I don't know why. Your answer and explanation will be greatly appreciated.
Let $Q$ be the field of rational numbers. We know the equation $x^3-2=0$ has three solutions $\sqrt[3]{2}$, $\sqrt[3]{2} \omega$, and $\sqrt[3]{2} \omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$. As such, the equation $x^3-4=0$ also has three solutions $\sqrt[3]{4}$, $\sqrt[3]{4} \omega$, and $\sqrt[3]{4} \omega^2$. None of these solutions belong to $Q$.
Now we do field extensions $Q\left ( \sqrt[3]{4} \right )$ and $Q\left ( \sqrt[3]{4} \omega \right )$. Because $\sqrt[3]{4}$ and $\sqrt[3]{4} \omega$ are both the solutions to $x^3-4=0$, we know that
$$ Q\left ( \sqrt[3]{4} \right ) \simeq Q\left ( \sqrt[3]{4} \omega \right ) .$$
where $\simeq$ represents 1-isomorphism. We then continue to extend both fields by $\sqrt[3]{2}$, we have
$$ Q\left ( \sqrt[3]{4},\sqrt[3]{2} \right ) \simeq Q\left ( \sqrt[3]{4} \omega,\sqrt[3]{2} \right ). $$
Because $\sqrt[3]{4} = (\sqrt[3]{2})^2$, we also have $$\sqrt[3]{4} \omega= (\sqrt[3]{2})^2,$$
due to the isomorphism.
I know the conclusion is wrong, but I don't know what was wrong during the deduction.
$\require{AMScd}$ Great question! Essentially, an abstract isomorphism of fields $K\simeq L$ need not be preserved when adjoining an element of a common extension field, because you need to make things compatible with the embeddings of $K$ and $L$ into the larger extension field. In your case, the element you want to adjoin already belongs to one of the fields, but not the other. So the first field will be unchanged, but the second will not.
What you're doing here is starting with an abstract isomorphism of fields $\sigma : \Bbb{Q}[\sqrt[3]{4}]\to\Bbb{Q}[\sqrt[3]{4}\omega]$, and then embedding both inside a larger extension field $\overline{\Bbb{Q}}$ which contains both. However, you cannot simply adjoin another element of $\overline{\Bbb{Q}}$ to both and expect them to remain isomorphic, because you've embedded these isomorphic fields inside $\overline{\Bbb{Q}}$ in different ways. Indeed, as you've observed, $\Bbb{Q}[\sqrt[3]{4},\sqrt[3]{2}] = \Bbb{Q}[\sqrt[3]{2}]$ (but $\Bbb{Q}[\sqrt[3]{4}\omega,\sqrt[3]{2}] = \Bbb{Q}[\omega,\sqrt[3]{2}]\not\simeq\Bbb{Q}[\sqrt[3]{2}]$).
The way to fix this is the following. The isomorphism $\sigma : \Bbb{Q}[\sqrt[3]{4}]\to\Bbb{Q}[\sqrt[3]{4}\omega]$ can be extended to an isomorphism $\tilde{\sigma} : \overline{\Bbb{Q}}\to \overline{\Bbb{Q}}.$ If you want to adjoin an element to both and preserve the isomorphism, you need to twist this element via the isomorphism $\tilde{\sigma}.$ That is, it will be true that $$ \Bbb{Q}[\sqrt[3]{4},\sqrt[3]{2}]\simeq\Bbb{Q}[\tilde{\sigma}(\sqrt[3]{4}),\tilde{\sigma}(\sqrt[3]{2})] = \Bbb{Q}[\sqrt[3]{4}\omega,\sqrt[3]{2}\omega^2]. $$ But if you omit this twist, all bets are off! The fields you get may no longer be isomorphic, as in your example. They also could be isomorphic! For example, if $\tilde{\sigma}(\alpha) = \alpha,$ then you will have $$\Bbb{Q}[\sqrt[3]{4},\alpha] \simeq\Bbb{Q}[\sqrt[3]{4}\omega,\alpha].$$
A very concise way to put all of this is that the diagram \begin{CD} \overline{\Bbb{Q}} @>\tilde{\sigma}>> \overline{\Bbb{Q}} \\ @AAA @AAA\\ \Bbb{Q}[\sqrt[3]{4}] @>\sigma>> \Bbb{Q}[\sqrt[3]{4}\omega]\\ @AAA @AAA\\ \Bbb{Q}@>\operatorname{id}>>\Bbb{Q} \end{CD} commutes, but the diagram \begin{CD} \overline{\Bbb{Q}} @>\operatorname{id}>> \overline{\Bbb{Q}} \\ @AAA @AAA\\ \Bbb{Q}[\sqrt[3]{4}] @>\sigma>> \Bbb{Q}[\sqrt[3]{4}\omega]\\ @AAA @AAA\\ \Bbb{Q}@>\operatorname{id}>>\Bbb{Q} \end{CD} does not.
Another way to think about what's going on here is that the polynomial $x^3 - 4$ is no longer irreducible over the field $\Bbb{Q}[\sqrt[3]{2}],$ so when you want to add a root of this polynomial to $\Bbb{Q}[\sqrt[3]{2}],$ the field you get will change depending on which root you picked.
Notice that in $\Bbb{Q}[\sqrt[3]{2}][x],$ we have $$ x^3 - 4 = (x - \sqrt[3]{2}^2)(x^2 + \sqrt[3]{2}^2x + \sqrt[3]{2}^4), $$ because over an algebraic closure of $\Bbb{Q},$ we have \begin{align*} x^3 - 4 &= (x - \sqrt[3]{2}^2)(x - \sqrt[3]{2}^2\omega)(x - \sqrt[3]{2}^2\omega^2)\\ &= (x - \sqrt[3]{2}^2)(x^2 - \sqrt[3]{2}^2\omega x - \sqrt[3]{2}^2\omega^2 x + (\sqrt[3]{2}^2)^2\omega^3)\\ &= (x - \sqrt[3]{2}^2)(x^2 - \sqrt[3]{2}^2x(\omega + \omega^2) + \sqrt[3]{2}^4)\\ &= (x - \sqrt[3]{2}^2)(x^2 + \sqrt[3]{2}^2x + \sqrt[3]{2}^4), \end{align*} as $\omega^2 + \omega + 1 = 0.$
Thus, we cannot simply adjoin a root of $x^3 - 4$ to the field $\Bbb{Q}[\sqrt[3]{2}]$. We instead have the isomorphism $$\Bbb{Q}[\sqrt[3]{2}][x]/(x^3 - 4)\cong\Bbb{Q}[\sqrt[3]{2}]\times\Bbb{Q}[\sqrt[3]{2}]/(x^2 + \sqrt[3]{2}^2x + \sqrt[3]{2}^4),$$ so the choice of root you adjoin to $\Bbb{Q}[\sqrt[3]{2}]$ changes which extension field you get.
Edit: To answer the question in your comment, that you want to think about adjoining a root of $x^3 - 2$ to $\Bbb{Q}[\sqrt[3]{4}]$ and not vice versa, I would argue that this is the wrong way to think about the situation. You're starting with two different isomorphic fields $\Bbb{Q}[\sqrt[3]{4}]$ and $\Bbb{Q}[\sqrt[3]{4}\omega]$ and then adjoining an element of a common algebraic closure, but there's no reason for these to remain isomorphic, because your abstract isomorphic between the two is not compatible with their embeddings into $\overline{\Bbb{Q}}$ (without twisting by the extension of your original isomorphism, anyway). This was what I was explaining in the first 3 paragraphs of my answer.
On the other hand, you can start with the same field ($\Bbb{Q}[\sqrt[3]{2}]$) and then try to adjoin different roots of the same polynomial, and this gives you different field extensions for the reason I've explained above. Since adjoining elements can be done in either order, I find this to be more of a philosophically satisfying way to think about the situation.
However, if you really do want to think about starting with $\Bbb{Q}[\sqrt[3]{4}]$ and then adjoining a root of $x^3 - 2,$ it turns out that again, $x^3 - 2$ is not irreducible in $\Bbb{Q}[\sqrt[3]{4}][x]$! In fact, we have \begin{align*} \frac{\sqrt[3]{4}^2}{2} &= \frac{\sqrt[3]{16}}{2}\\ &= \sqrt[3]{\frac{16}{8}}\\ &= \sqrt[3]{2}, \end{align*} and of course $\sqrt[3]{2}^2 = \sqrt[3]{4}.$ Together, these imply that $\Bbb{Q}[\sqrt[3]{4}] = \Bbb{Q}[\sqrt[3]{2}].$ Similarly, $\Bbb{Q}[\sqrt[3]{4}\omega] = \Bbb{Q}[\sqrt[3]{2}\omega^2],$ so both fields already have a particular root of $x^3 - 2.$ But, each field contains a different root of $x^3 - 2,$ so while on the one hand we have \begin{align*} \Bbb{Q}[\sqrt[3]{4}][x]/(x^3 - 2)&\cong\Bbb{Q}[\sqrt[3]{4}][x]/(x - \sqrt[3]{2})\times\Bbb{Q}[\sqrt[3]{4}][x]/(x^2+\sqrt[3]{2}x + \sqrt[3]{4})\\ &\cong\Bbb{Q}[\sqrt[3]{4}]\times\Bbb{Q}[\sqrt[3]{4}][x]/(x^2+\sqrt[3]{2}x + \sqrt[3]{4}), \end{align*} on the other hand we have \begin{align*} \Bbb{Q}[\sqrt[3]{4}\omega][x]/(x^3 - 2)&\cong\Bbb{Q}[\sqrt[3]{4}\omega][x]/(x - \sqrt[3]{2}\omega^2)\times\Bbb{Q}[\sqrt[3]{4}\omega][x]/(x^2 + \sqrt[3]{2}\omega^2 x + \sqrt[3]{4}\omega)\\ &\cong \Bbb{Q}[\sqrt[3]{4}\omega]\times\Bbb{Q}[\sqrt[3]{4}\omega][x]/(x^2 + \sqrt[3]{2}\omega^2 x + \sqrt[3]{4}\omega). \end{align*} The rings $\Bbb{Q}[\sqrt[3]{4}][x]/(x^3 - 2)$ and $\Bbb{Q}[\sqrt[3]{4}\omega][x]/(x^3 - 2)$ are of course abstractly isomorphic, but the two factors of the direct product decomposition are not.