In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.

With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.

For example, if $F=\mathbb F_p(x,y)$ and $K=\mathbb F_{p^2}(x,y,(x+\alpha y)^{1/p})$ for some $\alpha\in\mathbb F_{p^2}\setminus\mathbb F_p$, then the perfect closure of $F$ in $K$ is just $F$ itself, and the "problem" is exactly that the element whose $p$th root we took is not in $F$.

In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.

With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.

In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.

With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.

For example, if $F=\mathbb F_p(x)$ and $K=\mathbb F_{p^2}((x+\alpha)^{1/p})$ for some $\alpha\in\mathbb F_{p^2}\setminus\mathbb F_p$, then the perfect closure of $F$ in $K$ is just $F$ itself, and the "problem" is exactly that the element whose $p$th root we took is not in $F$.

In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.

With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.

For example, if $F=\mathbb F_p(x,y)$ and $K=\mathbb F_{p^2}(x,y,(x+\alpha y)^{1/p})$ for some $\alpha\in\mathbb F_{p^2}\setminus\mathbb F_p$, then the perfect closure of $F$ in $K$ is just $F$ itself, and the "problem" is exactly that the element whose $p$th root we took is not in $F$.