Why can't the notations $\mathbb{Q}( \sqrt[4]{4})$ and $\mathbb{Q}( \sqrt[4]{-4})$ be used for fields obtained by formal adjunction of zeros of respectively $X^4-4, X^4+4$ to $\mathbb{Q}$.
I know $\sqrt[4]{4}$ is a formal zero, not a number, but I still don't understand exactly what's wrong with it.
The notation is ambiguous because it apparently intends a root of $X^4-4$, but this polynomial is not irreducible. It factors into the product of $X^2-2$ and $X^2+2$, both of which are irreducible. The corresponding extensions of $\mathbf{Q}$ are $\mathbf{Q}(\sqrt{2})=\mathbf{Q}[X]/(X^2-2)$ and $\mathbf{Q}(\sqrt{-2})=\mathbf{Q}[X]/(X^2+2)$. These fields are not isomorphic.
In concrete terms, if you consider the subfields of $\mathbf{C}$ obtained by adjoining the $4$ roots of $X^4-4$, then you get two fields which are not isomorphic. Which one should $\mathbf{Q}(\sqrt[4]{4})$ mean?
The notation $\mathbf{Q}(\alpha)$ really only makes sense if $\alpha$ is an element of a given extension of $\mathbf{Q}$ (in which case it is the smallest subfield of that extension containing $\alpha$) or if $\alpha$ is meant to denote a root of an irreducible polynomial $f(X)\in\mathbf{Q}[X]$. Then the field $\mathbf{Q}(\alpha)$ admits the intrinsic description $\mathbf{Q}[X]/(f(X))$.
The issue is that if you replicate the formal process used when you obtain a field by writing $\mathbb{Q}(\sqrt[3]{2})$, in the cases you mention the result is not a field. This is a consequence of the fact that the polynomials suggested by the notations $\sqrt[4]{4}$ and $\sqrt[4]{-4}$ aren't irreducible (i.e. have nontrivial factorizations), as mentioned in the comments. If instead you take $\sqrt[4]{4}$ to refer to a particular complex number such as $\sqrt{2}$ or $\sqrt{2}i$, then the problem is that the resulting field depends on which fourth root of $4$ you choose, so the notation is ambiguous. This too results from the reducibility of the polynomials.
More precisely:
Option 1: by $\mathbb{Q}(\sqrt[4]{4})$, you intend to engage in a purely formal process. You adjoin to $\mathbb{Q}$ an element $\alpha$ satisfying the formal relation $\alpha^4 = 4$. Arithmetic in this new system should be done as normal, using the normal properties of $+,\times$, and simplifying using the relation $\alpha^4=4$ whenever possible.
This leads to a coherent arithmetic system (more precisely a ring), but it's not a field. You can see this for example by the fact that the elements $\alpha^2+2$ and $\alpha^2-2$ have product zero:
$$(\alpha^2+2)(\alpha^2-2) = \alpha^4 - 4 = 4-4 = 0$$
This makes it impossible for the elements $\alpha^2+2$ and $\alpha^2-2$ to have inverses. For example, if $\alpha^2+2$ had an inverse $\beta$, we'd have the following contradiction:
$$ \alpha^2 - 2 = (\alpha^2-2)\cdot 1 = (\alpha^2-2)\cdot (\alpha^2+2)\beta = 0\beta = 0$$
Option 2: by $\mathbb{Q}(\sqrt[4]{4})$ you intend to refer to the field generated over $\mathbb{Q}$ by one of the four complex numbers that are fourth roots of $4$, namely $\sqrt{2},-\sqrt{2},\sqrt{2}i,-\sqrt{2}i$. The problem here is that you get different (and not even isomorphic) fields depending on which choice you make. For example, in $\mathbb{Q}(\sqrt{2})$, $-2$ is not a square, but in $\mathbb{Q}(\sqrt{2}i)$, $-2$ is a square.
Remark: The process of formally adjoining an element $\alpha$ to a field (say $\mathbb{Q}$) that satisfies a certain algebraic relation $f(\alpha)=0$ is made precise with the notion of the quotient ring of a polynomial ring. The result is the quotient of the polynomial ring $\mathbb{Q}[x]$ by the principal ideal generated by $f$, i.e. $\mathbb{Q}[x]/(f)$. This is what's going on in Option 1 above. It is a good exercise to prove that the resulting ring is a field if and only if $f$ is irreducible.
$\frac{\mathbb{Q}\left[ X\right]}{\left(X^4 - 4 \right)} $ and $\frac{\mathbb{Q}\left[ X\right]}{\left(X^4 + 4 \right)} $ are direct products of fields (semisimple rings).
