So you’re asking for a crash course on separable extensions. It would go something like
this : Let us work in an algebraic closure $C$ of $K$.
Lemma 1 Let $P\in K[X]$ be an irreducible polynomial. Then $P$ is not separable
iff $P'=0$.
Proof of lemma 1 Suppose $P$ is inseparable, then there is $\alpha \in C$ such that
$\alpha$ is a multiple root of $P$, so $(X-\alpha)^2$ divides $P$. Writing $P=(X-\alpha)^2Q$ and differentiating, we see that $X-\alpha$ divides $P'$. If we call $G$
the gcd of $P$ and $P'$, then $G$ annihilates $\alpha$. Since $P$ is irreducible, we must
have $G=0$ and this is only possible if $P'=0$.
Conversely, suppose that $P'=0$. Let $\alpha$ be any root of $P$. We can write
$P=(X-\alpha)Q$, and hence $P'=Q+(X-\alpha)Q'$. We deduce $Q(\alpha)=0$, so $\alpha$ is amultiple root of $P$, so $P$ is not separable. QED
Corollary of lemma 1 In characteristic zero, every irreducible polynomial is separable.
We henceforth assume that we are in characteristic $p > 0$. We denote ${\sf Hom}_K(L,C)$
the set of all field homomorphisms $L \to C$ that fix pointwise every element of $K$.
Lemma 2 One has $|{\sf Hom}_K(K(\alpha),C)| \leq [K(\alpha):K]$, with equality
iff $\alpha$ is separable over $K$.
Proof of lemma 2 Any $\sigma \in {\sf Hom}_K(K(\alpha),C)$ is uniquely defined
by its value at $\alpha$, and this value must be a conjugate of $\alpha$, which yields the claimed inequality. Conversely, for each conjugate $\bar{\alpha}$ of $\alpha$, there is a unique $\sigma \in {\sf Hom}_K(K(\alpha),C)$ sending $\alpha$ to $\bar{\alpha}$, QED.
Corollary 1 of lemma 2 If we have a tower of extensions $L_1/L_2/\ldots /L_r$
and each $L_i$ is $L_{i-1}(\alpha_i)$ for some $\alpha_i$, then $|{\sf Hom}_{L_r}(L_1,C)| \leq [L_1:L_r]$, with equality iff each $\alpha_i$ is separable over $L_{i-1}$.
Corollary 2 (of Corollary 1 above) If we have an extension $L/K$ of finite degree,
then $|{\sf Hom}_K(L,C)| \leq [L:K]$, with equality iff $L$ is separable over $K$.
All the properties you need follow from this last corollary.