I first restate just the parts pertaining to your first question: if $X$ and $Y$ are sets, and $F \subseteq X \times Y$ is a relation with the properties one wants of a function (namely: for all $x \in X$ there is $y \in Y$ with $(x,y) \in F$, and for all $x \in X$ and $y_1, y_2 \in Y$ we have that $(x,y_1) \in F$ and $(x, y_2) \in F$ implies $y_1 = y_2$), some people do indeed view the triple $f = (X,Y,F)$ as being the the function, and define a function to be such a triple.
You then ask: "But isn't a function itself a relation and therefore musn't we write $(X,Y,f)$ as the set in which the function acts?" The answer to the first part of this question--- "isn't a function itself a relation"--- meaning, isn't the function itself the set $F$--- is no (and hence the second half of the question is dispensed with entirely). In the definition you have stated, a function is not a relation, but an ordered triple whose entries are a set $X$, then another set $Y$, and then finally a relation between $X$ and $Y$ (ie, a subset of $X \times Y$) with the properties one wants of a function.
To get a sense of the purpose of this: think of the notion of equality of functions that it produces: to say that $(X,Y,F)$ and $(X',Y',F')$ are equal is to say more than just $F = F'$, but also that $X=X'$ and $Y = Y'$. The purpose of this definition of a function is to make explicit the roles of the sets $X$ and $Y$ in the specification of a function. Note that if you are given a function $(X,Y,F)$, you can infer what $X$ is from $F$ alone ($F$ is a set of ordered pairs, and by the defining properties of a function, $X$ is nothing more than the set of first entries of elements of $F$) but you cannot infer what $Y$ is from $F$ alone.
An example might help: let $X = \{0,1\}$ and let $Y = X$ and let $Z$ denote the subset $\{0\}$ of $X$.
Let $f: X \to Z$ denote the only possible function (namely the one whose rule is given by $f(x) = 0$ for both possible values of $x$). Let $g: X \to Y$ denote the function given by $g(0) = 0$ and $g(1) = 0$. If you think about it for a moment you see that the rules of $f$ and $g$ are both encoded by the same set of ordered pairs, namely $F = \{(0,0), (1,0)\}$. The only difference between the two is that in one case we are regarding $Z$ as the codomain, and in the other we are regarding the larger set $Y$ as the codomain. With the above set-theoretic definition of "function", $f$ and $g$ are different functions, because one is the set $(X,X,F)$ and the other is the set $(X,Z,F)$, and these two ordered triples differ in their second coordinate, so they are different ordered triples, so the functions (by this definition) are different.
If you think of the set $F$ as "being" the function, then you cannot distinguish between $f$ and $g$ (and more generally, the concept of "codomain" becomes hazy). Of course, for many applications of the function concept there is no conceivable reason to distinguish between $f$ and $g$--- but most mathematicians, I think, would at least want the option of seeing $f$ and $g$ as different, and if you want to formalize functions within set theory, then, the definition above allows you to do so. (Some people do take as a definition that a function from $X$ to $Y$ is a subset of $X \times Y$ with the defining properties of a function; it is a perfectly good definition but it does not allow one to distinguish between $f$ and $g$ above, and the concept of codomain is not encoded in this definition, although if you are not formally minded, you might not notice.)
I want to reiterate: people generally do not think of functions as being ordered triples, consisting of the domain set, the codomain set, and then the relation giving the rule--- it's very common to think of the function as just being the rule. But if you want to formalize the notion, you quickly find that equating a function with the relation that determines its rule leaves a useful distinction (namely the concept of codomain, or the set a function is "to") out. So the purpose of the Bourbaki definition is to explicitly build it in: a function is not just a relation with the defining properties one wants of a function, but it is an explicit choice of domain and codomain together with this data. (As I remarked earlier, the domain could actually be inferred from the relation, so this is somewhat "redundant"--- but the object of this definition is merely to formalize a way of viewing functions within set theory, not to do so in a "minimal" way.)
Hopefully the rest of your questions are slightly cleared up by this.
Best Answer
I use $E \subseteq \binom{V}{2}$. Although, I have seen it used elsewhere, it's probably not a standard notation.