Proof that any extension that is Galois in the classical sense is also Hopf Galois The usual action of $G$ on $L$ is by automorphisms that fix $\mathcal{K}$. Explicitly, for any $g\in G, l,m\in L, k\in \mathcal{K}$
\begin{align*}
g(l+m)&=g(l)+g(m)\\
g(lm)&=g(l)g(n)\\
g(k)&=k
\end{align*}
This extends $\mathcal{K}$-linearly to an action of $\mathcal{K}G$ on $L$. Explicitly, for any $h=\Sigma_g k_g g\in \mathcal{K}G, l,m\in L, k\in \mathcal{K}$
\begin{align*}
h(l+m)=\Sigma_g k_g g(l+m)&=\Sigma_g k_g g(l)+\Sigma_g k_g g(m)=h(l)+h(m)\\
h(lm)=\Sigma_g k_g g(lm)&=\Sigma_g k_g g(l)\Sigma_g k_g g(n)\\
h(k)=\Sigma_g k_g g(k)&=\Sigma_g k_g k
\end{align*}
Then $L$ is a $\mathcal{K}G$-module. Moreover, $L$ is a $\mathcal{K}G$-module algebra as
$$
h(1_L)=\Sigma_g k_g g(1_L)=\Sigma_g k_g (1_L)=\epsilon(h)1_L
$$
Does this work to show that $L$ is a $\mathcal{K}G$-module algebra?
Also, I'm struggling to show bijectivity for the map $j$ in order to establish that the extension is Hopf-Galois where $j:L\otimes \mathcal{K}G\rightarrow \mathrm{End}_\mathcal{K}(L)$ is defined by
$$j(l\otimes h)(m)=(1,\mu)(l\otimes h)(m)=lh(m)$$ for any $l,m\in L, h=\Sigma_g k_g g\in \mathcal{K}G$
.
 A: First of all, the action of $\mathcal{K}G$ on the product $lm$ should be $$\sum_gk_gg(lm) = \sum_gk_gg(l)g(m)$$
for all $l,m \in L$ and $\sum_gk_gg \in \mathcal{K}G$.
Secondly, regarding the question, probably you are a bit familiar with Hopf algebra theory, so you know that $\mathcal{K}G$ is a Hopf algebra with its usual $\mathcal{K}$-group algebra structure and $\Delta,\varepsilon,S$ uniquely determined by
$$\Delta(g) = g \otimes g, \qquad \varepsilon(g)=1_\mathcal{K}, \qquad S(g) = g^{-1}.$$
And probably you are a bit familiar with category theory as well, so you already read somewhere that the "linearization" functor $\mathsf{Grp} \to \mathsf{Alg}_{\mathcal{K}}$ sending any group $G$ to the group algebra $\mathcal{K}G$ is left adjoint to the "units" functor $(-)^\times:\mathsf{Alg}_{\mathcal{K}} \to \mathsf{Grp}$ sending any $\mathcal{K}$-algebra to the group of its invertible elements. That is to say, we have a natural bijection
$$\mathsf{Grp}(G,A^\times) \cong \mathsf{Alg}_\mathcal{K}(\mathcal{K}G,A)$$
induced by $\mathcal{K}$-linearization of maps.
With these preliminary remarks at hand, your first observation corresponds to
$$\mathsf{Grp}(G,\mathrm{Aut}_\mathcal{K}(L)) = \mathsf{Grp}(G,\mathrm{End}_\mathcal{K}(L)^\times) \cong \mathsf{Alg}_\mathcal{K}(\mathcal{K}G,\mathrm{End}_\mathcal{K}(L)),$$
which explains why $L$ becomes a $\mathcal{K}G$-module. In addition, since, in fact, $G$ acts by field automorphisms, we have that
$$g(lm) = g(l)g(m) \qquad \text{and} \qquad g(1_L) = 1_L$$
for all $l,m \in L$ and $g \in G$, which makes of $L$ a $\mathcal{K}G$-module algebra. Recall, indeed, that for a $\Bbbk$-Hopf algebra $H$, an $H$-module algebra $A$ is a $\Bbbk$-algebra and an $H$-module at the same time in such a way that both multiplication $A \otimes A \to A$ and unit $\Bbbk \to A$ of $A$ are left $H$-linear maps, where the action of $H$ on $A \otimes A$ is the diagonal one
$$h \cdot (a  \otimes b) = \sum_{(h)} h_1a \otimes h_2b$$
for all $h \in H$, $a,b\in A$, and the action of $H$ on $\Bbbk$ is via restriction of scalars along $\varepsilon$
$$h \cdot k = \varepsilon(h)k$$
for all $h \in H$, $k \in \Bbbk$.
Concerning the map $j$, I suppose you would like to prove the bijectivity of the map $$j:L \otimes_\mathcal{K} \mathcal{K}G \to \mathrm{End}_\mathcal{K}(L), \qquad l \otimes_\mathcal{K}g \mapsto [m \mapsto lg(m)]$$
if $L/\mathcal{K}$ is a Galois extension.
To prove that it is injective, you may take advantage of
Dedekind's Lemma: For any group $G$ and field $E$, the set of $E$-valued characters of $G$ is linearly independent in the vector space ${\rm Fun}(G,E)$ of $E$-valued functions on $G$.
Any $g \in G$ is a field automorphism of $L$ that fixes $\mathcal{K}$. In particular, it can be seen as a group homomorphism $L^\times \to L$. Thus any vanishing linear combination
$$0 = l_1g_1 + \cdots + l_ng_n$$
in $\mathrm{End}_\mathcal{K}(L)$ with $l_i \in L$ is also a vanishing linear combination a characters $L^\times \to L$ in ${\rm Fun}(L^\times,L)$ and hence $l_i = 0$ for all $i = 1,\ldots,n$ by Dedekind.
Concerning surjectivity, pick any $f \in \mathrm{End}_\mathcal{K}(L)$. We know that $f$ is uniquely determined by its values on a basis $\mathcal{B} = \{e_1,\ldots,e_n\}$ of $L$ over $\mathcal{K}$. Since $[L:\mathcal{K}]=n<\infty$, we can consider the $L$-linear isomorphism
$$\mathrm{End}_\mathcal{K}(L) \to L^n, \qquad f \mapsto (f(e_1),\ldots,f(e_n)).$$
Thus $\dim_L(\mathrm{End}_\mathcal{K}(L)) = n = [L:\mathcal{K}]$. It follows that
$$\dim_L(L \otimes_\mathcal{K}\mathcal{K}G) = \dim_L(LG) = \dim_L(\mathrm{End}_\mathcal{K}(L))$$
and hence $j$ has to be surjective, too.
It may be interesting for you to prove that also the converse holds: if $j$ is an isomorphism, then $L/\mathcal{K}$ is Galois with Galois group $G$. You already have all the ingredients, it is just a matter of finding which statements hold always and which ones were related to the Galois hypothesis.
