1) Functions over finite sets may be equivalently thought of as vectors in the usual sense. Fix a finite set $E$, ordered arbitrarily as $E = \{ e_1, e_2, \ldots, e_m \}$. Then the set of functions $\mathcal F = \{ f : E \to \mathbb R \}$ is a real vector space of dimension $|E| = m$. In fact, this has the standard basis $\{ \delta_e \colon E \to \mathbb R \}_{e \in E}$ given by \delta_e(e') = \begin{cases} 1, &e' = e, \\ 0, &e' \ne e. \end{cases}
Based on this idea alone, we are justified in calling a function $f : E \to \mathbb R$ a vector. After all, what are members of a vector space called, if not vectors? However, to cement the idea even more, we go a bit further. Imagine stacking up the values of the function $f$ as a column vector: $ f = \left( \begin{array}{c} f(e_1) \\ f(e_2) \\ f(e_3) \\ \vdots \\ f(e_m) \end{array} \right). $ Then what we get is a correspondence (an isomorphism, in fact) between the function space $\mathcal F$ and the more familiar space of $m$-dimensional column vectors. This "vector notation" is often preferred over the function representation, since it is more compact. Moreover, this works seamlessly in conjunction with matrices (and matrices are obviously important in linear algebra).
2) An incidence vector $\chi^F$ is essentially the characteristic function of the set $F \subseteq E$ -- seen as a vector. The characteristic function $\chi^F$ is defined as $ \chi^F(e) = \begin{cases} 1, &e \in F, \\ 0, &e \notin F. \end{cases} $ Correspondingly, we can write $\chi^F$ as the $m$-vector with $1$ in the $i^{th}$ position if $e_i \in F$, and $0$ otherwise.
3) Given two vectors $a$ and $b$, you might have seen the inner product between them, defined as $ a^\mathrm T b = \sum_{i=1}^m a_i b_i. $ Unfolding this to the function representation, we can define the inner product of two functions $f, g : E \to \mathbb R$ to be $ \langle f, g \rangle = f^\mathrm T g = \sum_{e \in E} f(e) g(e). $ In the special case when $f = w$ and $g = \chi^F$, we get $ \langle w, \chi^F \rangle = w^\mathrm T \chi^F = \sum_{e \in E} w(e) \chi^F(e) = \sum_{e \in F} w(e), $ which is $w(F)$ by definition.