A counterexample
Take as ground field $F=\mathbb F_2(u,v)$ and consider the polynomial $f(X)=X^6+uvX^2+u\in F[X]$. This polynomial is irreducible by Eisenstein. Let $F\subset K$ be the extension obtained by adjoining a root $a$ of $f(X)$ to F, so that $K=F[a]$, $[K:F]=6$ and $f(a)=0$ .The element $a^2\in K$ has minimal polynomial over $F$ the separable (because of degree $3$) polynomial $g(X)=X^3+uvX+u$ .
From this follows that the separable closure of $F$ inside $K$ is $F^{sep}=F[a^2]$ and that the extension $F^{sep}=F[a^2] \subset K=F[a]$ is purely inseparable, as expected.

And now comes the surprise: there are no elements in $K$ purely inseparable over $F$ !(except the elements of $F$, of course) and so $K$ is not separable over the purely inseparable closure $F^{perf}=F$ of $F$. The proof that an element of $r\in K$ purely inseparable over $F$ belongs to $F$ is easy, once you realize that such an element must satisfy $r^2\in F$ . [Expand $r$ in the $F$-basis $a^i, 0\leq i\leq5$ and calculate $r^2$ (easy in characteristic $2$ !), remembering that $f(a)=0$.You will see that $r$ was already in $F$.]
(Edit: Kevin's comment and link have reminded that I had already given analogous examples in characteristic $p>2$, and completely forgotten about them! The two answers now cover all positive characteristics.)

A positive result
If $F\subset K$ is normal , then indeed $K$ is separable over $F^{perf}$, the extensions $F^{sep}$ and $F^{perf}$ are linearly disjoint over $F$ and the canonical map $F^{sep} \otimes_F F^{perf} \to K$ is an isomorphism. (As an aside observe that this is one of the rare cases where the tensor product of two fields is a field)

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A counterexample
Take as ground field $F=\mathbb F_2(u,v)$ and consider the polynomial $f(X)=X^6+uvX^2+u\in F[X]$. This polynomial is irreducible by Eisenstein. Let $F\subset K$ be the extension obtained by adjoining a root $a$ of $f(X)$ to F, so that $K=F[a]$, $[K:F]=6$ and $f(a)=0$ .The element $a^2\in K$ has minimal polynomial over $F$ the separable (because of degree $3$) polynomial $g(X)=X^3+uvX+u$ .
From this follows that the separable closure of $F$ inside $K$ is $F^{sep}=F[a^2]$ and that the extension $F^{sep}=F[a^2] \subset K=F[a]$ is purely inseparable, as expected.

And now comes the surprise: there are no elements in $K$ purely inseparable over $F$ !(except the elements of $F$, of course) and so $K$ is not separable over the purely inseparable closure $F^{perf}=F$ of $F$. The proof that an element of $r\in K$ purely inseparable over $F$ belongs to $F$ is easy, once you realize that such an element must satisfy $r^2\in F$ . [Expand $r$ in the $F$-basis $a^i, 0\leq i\leq5$ and calculate $r^2$ (easy in characteristic $2$ !), remembering that $f(a)=0$.You will see that $r$ was already in $F$.]

A positive result
If $F\subset K$ is normal , then indeed $K$ is separable over $F^{perf}$, the extensions $F^{sep}$ and $F^{perf}$ are linearly disjoint over $F$ and the canonical map $F^{sep} \otimes_F F^{perf} \to K$ is an isomorphism. (As an aside observe that this is one of the rare cases where the tensor product of two fields is a field)