Let's start with a useful proposition from a different text:
Proposition 21. (p.529 Dummit and Foote's "Abstract Algebra"):
Let $K_1$ and $K_2$ be two finite extensions of a field $F$ contained in $K$. Then:
$$[K_1K_2:F]\leq [K_1:F][K_2:F]$$
with equality iff an $F$-basis for one of the fields remains linearly independent over the other field. If $\alpha_1,...,\alpha_n$ and $\beta_1,...,\beta_m$ are bases for $K_1$ and $K_2$ over $F$, respectively, then the elements $\{\alpha_i\beta_j\}_{i,j}$ span $K_1K_2$ over $F$.
Immediately, we see that if $E/K$ and $E'/K$ are finite extensions, so is $E\cdot E'/K$. $\square$
Moreover, this gives us the form of the composite extension:
$$E\cdot E'/K = K(\{\alpha_i\beta_j\}_{i,j}),$$
where the $\alpha_i$'s and $\beta_j$'s are the respective algebraic elements over $K$ from $E$ and $E'$.
By definition (p.562), $K/F$ is Galois when $|Aut(K/F)| = [K:F]$.
In general, for a composite field:
$$|Aut(E/K)|\cdot|Aut(E'/K)|\leq|Aut(E\cdot E'/K)|$$
$\color{orange}{\text{[Pf: Exercise]}}$. Also, in general:
$$|Aut(K/F)|\leq [K:F]$$
$\color{orange}{\text{[Pf: Exercise]}}$. Combining these results, we get in the event that $E/K$ and $E'/K$ are Galois:
$$|Aut(E/K)||Aut(E'/K)| \leq |Aut(E\cdot E'/K)|\leq [E\cdot E':K]\leq [E:K][E':K] = |Aut(E/K)||Aut(E'/K)|.$$
This squeezes together $|Aut(E\cdot E'/K)| = [E\cdot E':K]$, making the composite Galois. $\square$
Part (i): Show the homomorphism:
$$\varphi:Gal(E\cdot E'/E) \to Gal(E'/E\cap E');$$
$$\sigma \mapsto \sigma|_{E'}$$
is an isomorphism.
Proof:
$\underline{\text{1-1:}}$ Suppose $\sigma,\tau\in Gal(E\cdot E'/E)$ are such that $\varphi(\sigma) = \varphi(\tau)$. Then:
$$\sigma|_{E'} \equiv \tau|_{E'}.$$
As these fix $E$, we know $\sigma|_E \equiv Id_E \equiv \tau|_E$. Lastly, since they are automorphisms they agree over linear combinations (constructing the composite field). So $\sigma = \tau$.
$\underline{\text{Onto:}}$ Take $\sigma \in Gal(E'/E\cap E')$, then trivially (and linearly) extend to $\widetilde{\sigma}\in Gal(E\cdot E'/E)$ by:
$$\widetilde{\sigma}|_E \equiv Id_E\text{ }\text{ }\text{ and }\text{ }\text{ }\widetilde{\sigma}|_{E'}\equiv \sigma|_{E'}.$$
Then we have existence of $\widetilde{\sigma}$ such that $\varphi(\widetilde{\sigma}) = \widetilde{\sigma}|_{E'} \equiv \sigma|_{E'} = \sigma$. $\square$
Part (ii): Show the homomorphism:
$$\psi:Gal(E\cdot E'/K)\to Gal(E/K)\times Gal(E'/K);$$
$$\sigma \mapsto (\sigma|_E,\sigma|_{E'})$$
is injective. In addition, if $E\cap E' = K$, then $\psi$ is surjective and hence an isomorphism.
Proof:
$\underline{\text{1-1:}}$ Just because, suppose $\sigma \in ker(\psi)$, then $\psi(\sigma) = (\sigma|_E,\sigma|_{E'}) = (Id|_E,Id|_{E'})$. Because $\sigma$ is an automorphism on $E\cdot E'/K$, we can separate the $K$-linear combinations, where all images (including $K$) are fixed. Hence $\sigma = Id_{E\cdot E'/K}$. In other words, $ker(\psi)$ is trivial.
$\underline{\text{Onto:}}$ Suppose $(\sigma,\tau)\in Gal(E/K)\times Gal(E'/K)$. Define $\widetilde{\sigma}\in Gal(E\cdot E'/K)$ by:
$$\widetilde{\sigma}: e\cdot e' \mapsto \sigma(e)\tau(e')$$
and extend $K$-linearly. We get $\psi(\widetilde{\sigma}) = (\widetilde{\sigma}|_E,\widetilde{\sigma}|_{E'}) = (\sigma,\tau).$ This seems to hold regardless of what $K$ is. $\blacksquare$
In closing, we really only required the extensions $E/K$ and $E'/K$ to be Galois to make the composite field Galois (see the second to last equality around the second '$\square$'). To me, this whole thing could be glossed over by just writing $Aut$...
