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 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 $L^H$.