Examples of Galois Correspondence being inapplicable with non-normal groups and fields

You're not going to find a counterexample inside a Galois extension, no matter what subgroup or intermediate field is used. The Galois correspondence holds for all intermediate fields and all subgroups (normal or not normal) in a Galois extension. It does not work when the whole field extension is not Galois.

For example, take $K = \mathbf Q$ and $L = \mathbf Q(\sqrt[4]{2})$. Then $L/K$ is not a normal extension ($x^4 - 2$ is irreducible over $K$ with degree $4$ and has a root in $L$ without splitting completely over $L$) and the number of $K$-automorphisms of $L$ is $2$ rather than $4$, which is the degree of the extension. Explicitly, $$ {\rm Aut}(L/K) = \{\sqrt[4]{2} \mapsto \sqrt[4]{2}, \sqrt[4]{2} \mapsto -\sqrt[4]{2}\}. $$ This group has fixed field $\mathbf Q(\sqrt{2})$, so $$ \Phi(\Gamma(K)) = \Phi(\Gamma(\mathbf Q)) = \Phi({\rm Aut}(L/K)) = \mathbf Q(\sqrt{2}) \not= K. $$

You're not going to find a counterexample for the other realtion, $\Gamma(\Phi(H)) = H$. Artin showed that if $H$ is a finite group of field automorphisms of a field $L$ then $L$ is always a finite Galois extension of $\Phi(H)$ (the subfield of $L$ fixed by $H$) and $\Gamma(\Phi(H)) = {\rm Gal}(L/\Phi(H)) = H$.

So the problem always occurs if you start with a finite extension of fields that is "bad" (not a Galois extension) rather than with a field and a finite group of automorphisms of it. If $L/K$ is a finite extension of fields and not a Galois extension, the subfield of $L$ fixed by ${\rm Aut}(L/K)$ is always larger than $K$. The example $\mathbf Q(\sqrt[4]{2})/\mathbf Q$ is a basic one that shows how more elements than the bottom field can be fixed by all the automorphisms of the top field that fix the bottom field. This illustrates what can go wrong if the extension is not a normal extension. What it does not show (and only a proof can really show it) is that the same problem happens every time the field extension is not normal. Also the same problem happens every time the field extension is not separable. That is, a finite extension $L/K$ is Galois if and only if $\Phi(\Gamma(K)) = K$.

For the record, I find the $\Phi$ and $\Gamma$ notation from your textbook very strange. I know $\Phi$ is a reminder of "field" and $\Gamma$ is a reminder of "group", but it is highly nonstandard in my experience. In most discussions of Galois theory, $\Gamma(E)$ is written as ${\rm Gal}(L/E)$ if you know $L/K$ is Galois (or perhaps ${\rm Aut}(L/E)$ if you don't yet know whether $L/K$ is Galois) and $\Phi(H)$ is written as $L^H$.

Through his unintuitive and idiosyncratic notation, Howie is trying to explain that if the subgroup isn't normal as a subgroup of the Galois group of the splitting field over the rational numbers, then the fixed field of that subgroup isn't normal as a field extension of the rational numbers.

Background: Let $F = \Bbb{Q}( \sqrt[4]{2}, i )$ be the splitting field of $x^4 - 2$ over $\Bbb{Q}$. For visualization purposes, it's most natural to understand $\operatorname{Gal}(F, \Bbb{Q})$, the group of $\Bbb{Q}$-preserving field automorphisms of $F$, as being generated by two elements $a, b$, where $a$ maps $\sqrt[4]{2}$ to $i \sqrt[4]{2}$, but fixes $\pm i$, and $b$ is complex conjugation. Clearly $a$ permutes the four roots of $x^4 - 2$ via $90°$ rotation CCW in the complex plane, and $b$ is reflection in the $x$-axis, which analogizes nicely with our ordinary geometric understanding of $D_4$ (the symmetry group of a square).
Having named these two elements, and understanding the expression $xy$ as performing transformation $y$, followed by transformation $x$, we find that $$D_4 = \{ e, a, a^2, a^3, b, ab, a^2 b, a^3 b \}.$$ In this notation, Howie's non-normal order $2$ subgroup $H_7$ (why $7$? Was he just numbering the subgroups of $D_4$ randomly?) is then $$H_7 = \langle a^2 b \rangle,$$ while the normal subgroup of order $2$ is $\langle a^2 \rangle$. Here $\langle x_1, ..., x_n \rangle$ indicates the subgroup generated by the elements $x_1, ..., x_n$.

The fixed field of the normal subgroup $\langle a^2 \rangle$ is $\Phi( \langle a^2 \rangle ) = \Bbb{Q}(i, \sqrt{2})$, which is indeed a normal extension, as it's the splitting field of $x^4 - 4$ over $\Bbb{Q}$. The Galois group $\operatorname{Gal}(\Bbb{Q}(i, \sqrt{2}) : \Bbb{Q})$ is the Klein four-group $$V = \{e, x, y, xy \},$$ where $x$ swaps $i$ with $-i$ and $y$ swaps $\sqrt{2}$ with $-\sqrt{2}$. $V$ is commutative and each element is self-inverse, so the subgroups $\langle x \rangle$, $\langle y \rangle$, $\langle xy \rangle$ are all normal. These subgroups correspond to the subfields $\Bbb{Q}(\sqrt{2})$, $\Bbb{Q}(i)$, $\Bbb{Q}(i \sqrt{2})$. We can directly find that \begin{align*} \Gamma(\Bbb{Q}) = V &\text{ and } \Phi(V) = \Bbb{Q}, \\ \Gamma(\Bbb{Q}(\sqrt{2}) = \langle x \rangle &\text{ and } \Phi(\langle x \rangle) = \Bbb{Q}(\sqrt{2}), \\ \Gamma(\Bbb{Q}(i) = \langle y \rangle &\text{ and } \Phi(\langle y \rangle) = \Bbb{Q}(i), \\ \Gamma(\Bbb{Q}(i\sqrt{2}) = \langle xy \rangle &\text{ and } \Phi(\langle xy \rangle) = \Bbb{Q}(i\sqrt{2}), \\ \Gamma(\Bbb{Q}(\sqrt{2}, i)) = \{ e \} &\text{ and } \Phi(\{ e \}) = \Bbb{Q}(\sqrt{2}, i). \ \end{align*} In other words, because the subgroup was normal, the corresponding field extension is normal as well.

By contrast, the fixed field of $H_7$, in Howie's notation, is $\Phi(H_7) = \Bbb{Q}(i\sqrt[4]{2})$.
Just as the group $H_7 = \langle a^2 b \rangle$ is non-normal as a subgroup of $D_4 = \langle a, b \rangle$, the field $\Bbb{Q}(i\sqrt[4]{2})$ is non-normal as an extension of $\Bbb{Q}$.
Indeed, $x^4 - 2$ itself fails to split over $\Bbb{Q}(i\sqrt[4]{2})$: when we try to factor it, we get $$x^4 - 2 = (x - i\sqrt[4]{2})(x + i\sqrt[4]{2})(x^2 - 2),$$ and here we are stuck, because $\Bbb{Q}(i\sqrt[4]{2})$ contains a square root of $-2$, but no square root of $2$.

How can we use this to answer [part of] Howie's question?
Let $E = \Bbb{Q}$ considered as a subfield of the field extension $\Bbb{Q}(i\sqrt[4]{2})$.
The automorphism group $\operatorname{Aut}(\Bbb{Q}(i\sqrt[4]{2}):\Bbb{Q})$ has 2 elements $\{ e, g \}$: the identity $e$, and the map $g$ which swaps $i\sqrt[4]{2}$ with its negative.
Then $$\Gamma(E) = \Gamma(\Bbb{Q}) = \{ e, g \},$$ but $$\Phi(\Gamma(E)) = \Phi(\Gamma(\Bbb{Q})) = \Phi(\{ e, g \}) = \Bbb{Q}(\sqrt{2}),$$ which strictly contains $E = \Bbb{Q}$.

However, there are only two subgroups of $\{ e, g \}$: the trivial subgroup $\{ e \}$ and the entire group $\{ e, g \}$. We check that both of these subgroups satisfy $H = \Gamma(\Phi(H))$: \begin{align*} \Gamma(\Phi( \{ e \} )) = & \Gamma(\Bbb{Q}(i\sqrt[4]{2})) = \{ e \}, \\ \Gamma(\Phi( \{ e, g \} )) = & \Gamma(\Bbb{Q}(\sqrt{2})) = \{ e, g \}. \ \end{align*} So no subgroup $H < \operatorname{Aut}(\Bbb{Q}(i \sqrt[4]{2}): \Bbb{Q})$ can give us an example where $H$ is strictly contained in $\Gamma(\Phi(H))$.