Clarifying Notation (Sets & Functions) The basic problem is that I want to be extremely clear about the sets
that mathematical manipulations and operations are taking place in, I am
hoping for someone who really understands this to read what I've written
closely and point out what is getting me all mixed up, though of course
reading &/or responding isn't mandatory (lol) - but it is a long post even
though it's dealing with just one idea. 
The set-theoretic definition of a function is f = (X,Y,F) where F is a subset
of ordered pairs of the Cartesian product of X & Y, (i.e. F ⊆ (X x Y) a relation).
This is Bourbaki's way of defining a function and he (they) call F the graph.
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? To expand this out:
(X,Y,f) = (X,Y,(X,Y,F)).
I've come across notation that specifies (X,Y,f) as ((X,Y),f).
Here, page 35 of this .pdf file
So ((X,Y),f) = ((X,Y),((X,Y),F)) would seem to make sense.
Bourbaki calls f a set & F it's graph but the notation in the .pdf file says
that f would be defined in the way I've explained above, i.e. that F is a
subset of XxY. The thing is that since a function f is itself a relation
shouldn't it be a relation in a set, i.e. ((X,Y),f)?
Assuming that the above is the way to think about these things, how
would I think of both F & f? In f = (X,Y,F), F ⊆ (X x Y) so (x,y) ∈ F or xFy,
where obviously (x∈X) & (y∈Y).
How about f? I think f ⊆ (X x Y) so (x,y) ∈ f or xfy.
I don't understand how this makes sense because for the set f = (X,Y,F)
Bourbaki writes f : X → Y so for (X,Y,f) I'd have to set g = (X,Y,f) and
write g : X → Y. This is a weird conclusion but it seems to suggest itself.
The problem of being extremely clear about what sets you are using is
particularly interesting when doing linear algebra.
The use of set-theoretic notation in linear algebra both clarifies things for
me and brings up similar questions, for a vector space V I could write
((V,+),(F,+',°),•) with the clarification that:
in (V,+) we have + : V × V → V,
in (F,+',°) we have (+' : F × F → F) & ( ° : F × F → F).
In • we have (• : F × V → V) or perhaps [• : (V,+) × (F,+',°) → (V,+)]?
This notation clearly illustrates why the two operations, vector addition
and scalar multiplication are used on a vector space and the axioms for
each clearly jump out, i.e. (V,+) is abelian, (F,+',°) is a field and • isn't the
clearest to me but I think it's similar to the way that + & ° are related in a
field, i.e. "multiplication distributes over addition".
Relating all of this to the concerns I had above in a clear manner, in the
set (F,+',°) it would make sense that +' is a set of the form (F,+'') where
+'' is a subset of the cartesian product of F x F. Similarly with °, and in the
set (V,+) you'd have something similar, also in • you'd have a crazy set
((V,+), (F,+',°), •') or including even more brackets (((V,+), (F,+',°)), •')
with •' being a subset of the cartesian product of (V,+) & (F,+',°).
There is another problem when you want to give a vector space a norm,
would I write ((V,+),(F,+',°),•,⊗) where ⊗ : V x V → F ? Would ⊗ itself
suggest the subset ((V,+), (F,+',°), ⊗') in the manner explained above?
I don't think so because ⊗' would be the set of ordered pairs (x,a) with
x ∈ V and a ∈ F but since V x V → F you've got the map (x,x') ↦ a, it's
quite confusing tbh and need help with this.
All this seems crazy but it also makes a lot of sense, I want to be very
rigorous about what I'm doing and all of the above seems to suggest itself
but it could be a lot of nonsense caused by simple confusion of a particular
issue in the post , I'm thinking that (X,Y,F) implying (X,Y,f) is the
culprit but again this idea clarifies things. If you read to this point thanks
so much, hopefully you recognise the issue.
 A: 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.
