First, suppose that $L/K$ is separable. Fix an algebraic closure $\overline{K}$ for $K$, and let $M$ be a Galois closure for $L/K$ in $\overline{K}$.
EDIT: If the idea of Galois closure isn't known to you, think about it as follows. Since your extension is separable, it is primitive, say $L=K(\alpha)$. Note then that necessarily $K(\alpha')/K$ is separable for every root $\alpha'$ of the minimal polynomial $m_\alpha$ of $\alpha$ over $K$. In the algebraic closure $\overline{K}$, all the roots $\alpha'$ exist. Then, the splitting field $F$ of $m_\alpha$ is just $K(\{\alpha':m_\alpha(\alpha')=0\})$. Since each $\alpha'$ is separable the full extension $F$ is separable. What this basically shows is that every finite separable extension of $K$ is contained in a finite Galois extension of $K$ (in fact, if you fix an algebraic extension, this is the minimal Galois extension containing $K$). So, when I say "take a Galois closure" I just mean "think about a Galois extension containing our extension". This is a common technique, for it allows you to deduce results about separable extensions by first proving it for Galois extensions, and then deducing it for certain subextensions.
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Note that for any embedding $\tau\in\text{Hom}_K(L,\overline{K})$ one necessarily has that $\tau(L)\subseteq M$. Thus, for every $\sigma\in\text{Gal}(M/K)$ and every $\tau\in\text{Hom}_K(L,\overline{K})$ one has that $\sigma\circ\tau\in\text{Hom}_K(L,\overline{K})$. Moreover, since $\sigma\circ\tau=\sigma\circ\tau'$ implies that $\tau=\tau'$, you can see, in particular, that the mapping $\tau\mapsto \sigma\circ\tau$ is an injection $\text{Hom}_K(L,\overline{K})$ to itself, and so necessarily a bijection. From this you can see that for every $\sigma\in\text{Gal}(M/K)$ and $x\in L$ one has that
$$\begin{aligned}\sigma\left(\text{Tr}_{L/K}(x)\right) &= \sigma\left(\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\right)\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}(\sigma\circ\tau)(x)=\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\\ &= \text{Tr}_{L/K}(x)\end{aligned}$$
Since $M/K$ is Galois, this implies that $\text{Tr}_{L/K}(x)\in K$.
Do something similar for the norm map.
For your second question. Use the fact that $F\subseteq F(\alpha)\subseteq K$ and the fact that degree is multiplicative in towers. To deduce the other result, consider the polynomial
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(\alpha))$$
Convince yourself that for each distinct Galois conjugate $\beta$ of $\alpha$, that $\sigma(\alpha)=\beta$ precisely $\frac{n}{d}$ times. Thus,
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(x))=m_\alpha(T)^{\frac{n}{d}}$$
Compare the constant term and the coefficient of $T^{n-1}$ on both sides to get both of your results.
I’m going to interpret your question as asking whether $\mathrm{Aut}(K_1K_2/F) = \mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)$ if $K_1$ and $K_2$ are extensions of $F$ such that $K_1\cap K_2=F$, where for fields $L\subseteq M$,
$$\mathrm{Aut}(M/L) = \{\sigma\in\mathrm{Aut}(M)\mid \sigma|_L=\mathrm{id}_L\},$$
and hence equals the Galois group in the special case where $M$ is Galois over $L$.
Consider $F=\mathbb{Q}$, $K_1=\mathbb{Q}(\sqrt[3]{2})$, and $K_2=\mathbb{Q}(\omega)$, where $\omega$ is a complex cubic root of unity; that is, a root of $x^2+x+1$.
Then $K_2$ is Galois over $\mathbb{Q}$ and $\mathrm{Gal}(K_2/F)$ is cyclic of order $2$. On the other hand, $\mathrm{Aut}(K_1/F)$ is trivial, because $\sqrt[3]{2}$ is the only real cubic root of $2$, $K_1$ is contained in $\mathbb{R}$, and an automorphism of $K_1$ must send $\sqrt[3]{2}$ to a cubic root of $2$. Thus,
$$\mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)\cong\{e\}\times C_2 \cong C_2$$
is cyclic of order $2$.
However, $K_1K_2$ is the splitting field of $x^3-2$, which has degree $6$ over $F$. In particular, $\mathrm{Gal}(K_1K_2/F)$ has order $6$ (and is isomorphic to $S_3$) so we get neither an isomorphism with, nor a realization as a subgroup of the direct product.
Best Answer
We have the isomorphism $$Gal(K_1\cap K_2 / F) \cong Gal(K_2/F)\ /\ Gal(K_2 / K_1\cap K_2).$$ Hence, if $\sigma\in Gal(K_1/F)$, the restriction $\sigma_{\mid K_1\cap K_2}$ corresponds to a conjugacy class in said quotient, which has size $|Gal(K_2/K_1\cap K_2)|$.