The intuition I can provide is that in an algebraic extension $F\subset K$ of characteristic $p$ , the field $K$ will be a separable extension of its purely inseparable extension $K_p$ over $F$ whenever $trdeg_{\mathbb F_p} F\leq 1$. This is why my counterexamples on MathOverflow here and here have $F=\mathbb F_p(u,v)$ , a purely transcendental extension of $\mathbb F_p$ of degree two.
I can't prove my intuition but it implies that Zev's example won't work (he didn't say it did, so he made no mistake), and here is a proof that indeed it doesn't.
The key point is that is that if $\mathbb F_q$ is a finite field of characteristic $p$, the Frobenius map $\mathbb F_q \to \mathbb F_q: x\mapsto x^p$ is surjective and this implies that for $f(T)\in \mathbb F_q [T]$ we have
$ (f(T))^{1/p}\in \mathbb F_q [T^{1/p}]$, because of the freshman's dream
$(a+bT+\ldots)^{1/p}=a^{1/p}+b^{1/p}T^{1/p}+\ldots $ !
This implies that
$$(\mathbb F_q(T))^{p^{-\infty}}=\mathbb F_q(T,T^{p^{-1}},T^{p^{-2}},\ldots)$$
And now the rest is easy. If $F= \mathbb F_p(T)$ and
$K=(\mathbb F_{p^2}(T))^{p^{-\infty}}=\mathbb F_{p^2}(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$ we have
$K_p=\mathbb F_p(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$ and $K$ is a separable extension of the perfect closure $K_p$ of $F$ in $K$ . Indeed $K$ is just the simple separable extension $K=K_p(g)$ of $K_p$, where $g \in \mathbb F_{p^2}$ is any element such that $\mathbb F_{p^2}=\mathbb F_{p}(g)$. [ The element $g$ is of course separable over the (perfect !) field $\mathbb F_p$, so it is a fortiori separable over $K_p$].
Edit What I called my intuition above has now been proved by ulrich on MathOverflow here. He proves that if $trdeg_{\mathbb F_p} F\leq 1$, any algebraic extension $F\subset K$ has the property that $K$ is separable over its
purely inseparable extension $K_p$ over $F$.
$L/K(a_1,\ldots,a_n)$ is algebraic and $K(a_1,\ldots,a_n)/K$ is purely transcendental.
In general there is no other canonical (*) decomposition as shown with
$\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$ (an elliptic curve) whose field of constants is $\Bbb{Q}$.
(*) the decomposition depends on a choice of transcendental basis so it is not really canonical, it is interesting to check if there are some natural conditions giving a particular one. Minimizing $[L:K(a_1,\ldots,a_n)]$ is natural.
For your second question, the non-separable normal algebraic extensions decompose in the two ways: $K(L^{p^\infty})$ means $\bigcap_{n\ge 1} K(L^{p^n})$ where $p$ is the finite characteristic, then $L/K(L^{p^\infty})$ is purely inseparable and $K(L^{p^\infty})/K$ is separable, while with $H$ the $K$-embeddings $L\to \overline{L}$ and $L^H$ the subfield fixed by all the embeddings then $L/L^H$ is separable and $L^H/K$ is purely inseparable.
When $L/K$ is not normal then $L/L^H$ doesn't have to be separable as shown there but I didn't know it before so I don't have much to say about it.
Best Answer
Yes, the first statement is true. There is no contradiction with what you write: indeed, the strategy for the problem is to decompose the algebraic extension as a separable extension followed by a purely inseparable extension, note that the separable extension must be trivial, and then argue from there.
The answer to your second question is also yes: you can take the perfect closure of any field $k$. One way to do this is to embed $k$ in to an algebraic closure $\overline{k}$ and then take all the elements $x\in \overline{k}$ so that $x^{p^n}\in k$ for some $n$.