Number of embeddings in algebraic closure

The idea behind the proof is that for a field $K$ and an element $\alpha \in \bar{K}$, the roots of the minimal polynomial of $\alpha \in \bar{K}$ are exactly the conjugates of $\alpha$ over $K$. Then taking $L = K(\alpha)$ each conjugate of $\alpha$ defines a unique embedding from $L$ to $\bar{K}$. Since $[L: K] = n$, there are $n$ distinct embeddings.

For the full details of this proof look at Lemma 5.17 and Theorem 5.18 of this.

I'll provide a slightly different solution to your second question. Assuming that $\#\text{Hom}(L,\bar K)=n$, let $x=x_1,\dots,x_m\in \bar K$ be the conjugates of $x$ (i.e., the roots in $\bar K$ of the minimal polynomial $m_{x,K}(t)$ for $x$ over $K$). Recall that each of the $m$ isomorphisms $K(x)\rightarrow K(x_j)$ extend to $d$ isomorphisms $L=K(x,\theta)\rightarrow K(x_j,\theta_k)$ where $\theta\in \bar K$ is a primitive element for $L$ over $K(x)$ (i.e. $L=K(x,\theta)$) and $\theta=\theta_1,\dots,\theta_d$ are its conjugates. (This is Theorem 8 in section 13.1 of Dummit and Foote.) This accounts for all $n=md$ embeddings $L\rightarrow \bar K$ via $L=K(x,\theta)\rightarrow K(x_j,\theta_k)\rightarrow\bar K$, where the last map is inclusion. Thus, what we find is that for each $1\leq j \leq m$ there are $d$ embeddings $\sigma:L\rightarrow \bar K$ with $\sigma: x\mapsto x_j$. Hence there are $m$ equivalence classes, each with $d$ elements.