Let $A$ be some normed vector space and let $A^\ast$ denote its dual. Then if for all $\varphi \in A^\ast$ we have $\varphi (a) = 0$ then $a=0$. This is an argument that is frequently used.
Consider this theorem: If $A$ is a unital $C^\ast$ algebra and $a \in A$ is normal and $f \in C(\sigma (a)) $ and $g \in C(\sigma (f(a))$ then
$$ (g \circ f)(a) = g(f(a))$$
The proof (which I don't understand!) is this:
If $C$ denotes the $C^\ast$ algebra generated by $1$ and $f(a)$ then $c \subseteq B$ where $B$ is the algebra generated by $1$ and $a$ then $C \subset B$ and for any $\tau \in \Omega(B)$ its restriction $\tau_C$ is a character on $C$. We therefore have
$$ \tau ((g \circ f)(a)) = g(f(\tau (a))) = g(\tau_C(f(a))) = \tau_C(g(f(a)) = \tau( g(f(a)))$$
hence $(g\circ f)(a) = g(f(a))$.
It looks like the argument is that because for all $\tau \in \Omega(B)$ we have $\tau ((g \circ f)(a)) = \tau( g(f(a)))$ therefore $(g\circ f)(a) = g(f(a))$. But the characters $\Omega (B)$ is only a subset of the functionals $B^\ast$ so this must be a different argument used. What is going on here?
Also, why is $g(f(\tau (a))) = g(\tau_C(f(a)))$? It is not clear to me why $\tau$ can be changed to its restriction.
Regarding your first question, the characters separate points in a commutative C$^*$-algebra. That's a basic theorem that allows you to do the Gelfand transform.
Since $f(a)\in C$ and $\tau_C=\tau|_C$, we have $f(\tau(a))=\tau(f(a))=\tau_C(f(a))$. I agree that it is not obvious (to me, at least) why the need of using $\tau_C$.