Proof 1 Let $(a_1,\ldots, a_n)$ be a $k$ basis for $K/k$, then the following is the finite sequence of extensions. $$K\subset K(a_1)\subset K(a_1,a_2)\subset\ldots\subset KL $$

Proof 2 Since the extensions are finite the map $K\otimes_k L\ra KL$ is bijective, since $K\otimes_k L$ is finite dimensional so is $KL$.

Proof Intersection of two galois extensions of $k$ is visibly again galois and the uniqueness follows from it. To prove the existence choose a primitive element $x$ of $K/k$ (remember that $k$ is perfect). The field of roots of minimal polynomial of $x$ over $k$ is the required Galois extension (algebraic and contains all conjugates as well as $K$).

Proof 1. Let $x$ be a primitive element of $K/k$ (that is $K=k[x]$) with minimal polynomial $P\in k[X]$, hence $KL=L[x]$ and the minimal polynomial $Q$ of $x$ on $L$ divides $P$ so that the roots are in $K$ like those of $P$ (since $K/k$ is Galois) and a fortiori in $KL$. As $P$ has simple roots it proves that $KL/L$ is Galois. More precisely, $Q=\prod(X-x_i)$ where $x_i\in K$ are conjugates of $x$, and thus is also in $K[X]$, which proves that we have $Q\in (K\cap L)[X]$.
To prove the surjectivity of $r$. Let $g\in\gal(K/K\cap L)$, then $g(Q(x))=Q(g(x))$ because $Q$ has coefficients in $K\cap L$. By the universal property of quotient, there exists a unique $L$ endomorphism of $KL=L[X]/Q$ which sends $x=(X\mod Q)$ to $g(x)$, which is the sought antecedent.
To prove the injectivity of $r$. Let $g\in\gal(KL/L)$ be in the kernel, that to say trivial on $K$. Since $g$ is trivial on $L$ and $K,L$ generate $KL$, it is trivial on $KL$, what we wanted.
2. Suppose further that $L$ is Galois. Then, $L$ is the field of roots of a separable polynomial $P_1\in k[X]$ and $KL$ is the field of roots of separable polynomial $\lcm(P, P_1)$, proving that $KL/k$ is galois. For the latter point, let $\sigma\in\Hom_k(K\cap L,\Omega)$, which is extended to entire $KL$. As $K,L$ are galois on $k$, we have $\sigma(K)\subset K$ and $\sigma(L)\subset L$ and thus $\sigma(K\cap L)\subset K\cap L$, hence the equality (because $K \cap L$ is of finite dimension on $k$ and we conclude as usual).

Proof. The first part of the theorem gives $\gal(KL/L)\simeq \gal(K/K\cap L)$ giving the first equality.
For the second part $$[KL:k]=[KL:L][L:k]=[K:K\cap L][L:k]=\frac{[K:k]}{[K\cap L:k]} [L:k]$$ giving the second result.

The restriction morphisms $$\gal(K/k)\ra \gal(K\cap L/k)\quad\text{and}\quad\gal(L/k)\ra \gal(K\cap L/k) $$ define a morphism $$(j_1,j_2):\gal(K/k)\times \gal(L/k) \ra \gal(K\cap L/k)\times \gal(K\cap L/k)$$ There is also a natural injective map $i:\gal(KL/k)\ra\gal(K/k)\times \gal(L/k)$ which can be composed as $j_1\circ i$ and $j_2\circ i$ to give $$\gal(KL/k)\ra \gal(K\cap L/k) $$

Proof It suffices to show that the cardinality of two groups are equal. We have an exact sequence $$1\ra N\ra \gal(K/k)\times \gal(L/k)\xra{(j_1,j_2)} \gal(K\cap L/k)\times\gal(K\cap L/k)\ra 1 $$ Since $j_1,j_2$ are restriction maps they are surjective and thus also $(j_1,j_2)$. By construction, our amalgamated product say $G$, is the inverse image by $(j_1,j_2)$ of the diagonal subgroup $$\gal(K\cap L/k)=\{(g,g), g\in \gal(K\cap L/k)\}\subset \gal(K\cap L/k)\times\gal(K\cap L/k). $$ We thus have an exact sequence $$1\ra N\ra G\ra\gal(K\cap L/k)\ra 1 $$ Comparing the cardinality gives $$\card G= \frac{[K:k][L:k]}{[K\cap L:k]} =[KL:k]$$ where the last equality follows from Corollary 4. $\blacksquare$