How can an ultrafilter be considered as a finitely additive measure?

It does not say that every finitely additive measure induces an ultrafilter. However, it is true that every non-trivial $\{0,1\}$-valued finitely additive measure on $\wp(X)$ induces an ultrafilter on $X$.

Suppose that $m$ is a $\{0,1\}$-valued finitely additive measure defined on $\wp(X)$ such that $m(X)=1$. Let $\mathscr{U}=\{U\subseteq X:m(U)=1\}$. For each $A\subseteq X$ we have $m(A)+m(X\setminus A)=1$, so exactly one of $A$ and $X\setminus A$ belongs to $\mathscr{U}$. Clearly $V\in\mathscr{U}$ whenever $X\supseteq V\supseteq U\in\mathscr{U}$, so $\mathscr{U}$ is closed under taking supersets. If $U,V\in\mathscr{U}$ and $U\cap V\notin\mathscr{U}$, then

$$1=m(U)=m(U\setminus V)+m(U\cap V)=m(U\setminus V)$$

and similarly $m(V\setminus U=1$, so

$$m(U\cup V)=m(U\setminus V)+m(U\cap V)+m(V\setminus U)=2\;,$$

which is absurd. Thus, $\mathscr{U}$ is closed under taking finite intersections. Finally, for each $A\subseteq X$ we have $m(A)+m(X\setminus A)=1$, so exactly one of $A$ and $X\setminus A$ belongs to $\mathscr{U}$. Thus, $\mathscr{U}$ is an ultrafilter on $X$.

Yes, it is enough (if we insist that the whole set has measure $1$). We need to check that any superset of a set of measure $1$ has measure $1$ (easy), and that the intersection of two sets of measure $1$ has measure $1$.

So let $A$ and $B$ have measure $1$. Then $A\cup B$ has measure $1$, and is the disjoint union of $A\cap B$, $A\setminus B$, and $B\setminus A$.

If $A\cap B$ does not have measure $1$, then each of $A\setminus B$ and $B\setminus A$ do, contradicting finite additivity.

Remark: Finitely additive $\{0,1\}$-valued measures defined on all subsets of a set $I$, and ultrafilters on $I$ are such close relatives that there is no point in distinguishing between the two.

When we are working with an ultrapower $A^{I}/D$, it is often more natural to say that the functions $f$, $g$ from $I$ to $A$ are equal "almost everywhere" than to say $\{i:f(i)=g(i)\}\in D$.