Let me work in mathematicians' notation for a bit and then switch back to Dirac notation.
Suppose you start with a Hilbert space $\mathscr H$, which you can understand as a space of functions from some coordinate space $S$ into $\mathbb C$, i.e. if $f\in\mathscr H$ then $f:R\to \mathbb C$, and that you have some suitable notion of inner product $(·,·):\mathscr H\times \mathscr H\to\mathbb C$, like e.g. an integral over $R$. (Note that here $(·,·)$ should be linear on the second argument.)
Given this structure, for every vector $f\in\mathscr H$ you can define a linear functional $\varphi_f:\mathscr H\to \mathbb C$, i.e. a function tha takes elements $g\in \mathscr H$ and assigns them complex numbers $\varphi_f(g)\in \mathbb C$, whose action is given specifically by $\varphi_f(g) = (f,g)$. As such, $\varphi_f$ lives in $\mathscr H^*$, the dual of $\mathscr H$, which is the set of all (bounded and/or continuous) linear functionals from $\mathscr H$ to $\mathbb C$.
There's plenty of other interesting functionals around. For example, if $\mathscr H$ is a space of functions $f:R\to \mathbb C$, then another such functional is an evaluation at a given point $x\in R$: i.e. the map $\chi_x:\mathscr H\to\mathbb C$ given by
$$\chi_x(g) = g(x).$$
In general, this map is not actually bounded nor continuous (w.r.t. the topology of $\mathscr H$), but you can ignore that for now; most physicists do.
Thus, you have this big, roomy space of functionals $\mathscr H^*$, and you have this embedding of $\mathscr H$ into $\mathscr H^*$ given by $\varphi$. In general, though, $\varphi$ may or may not cover the entirety of $\mathscr H^*$.
The correspondence of this into Dirac notation goes as follows:
$f$ is denoted $|f\rangle$ and it's called a ket.
$\varphi_f$ is denoted $\langle f|$ and it's called a bra.
$\chi_x$ is denoted $\langle x|$, and it's also called a bra.
Putting these together you start getting some of the things you wanted:
2. $\langle x |f\rangle$ is $\chi_x(f) = f(x)$, i.e. just the wavefuntion.
6. $\langle f | g \rangle$ is $\varphi_f(g) = (f,g)$, i.e. the iner product of $f$ and $g$ on $\mathscr H$, as it should be.
Note in particular that these just follow from juxtaposing the corresponding interpretations of the relevant bras and kets.
7. Somewhat surprisingly, $\langle f | x\rangle$ is actually defined - it just evaluates to $f(x)^*$. This is essentially because, in physicists' brains,
9. $|x\rangle$ is actually defined. It's normally understood as "a function that is infinitelly localized at $x$", which of course takes a physicist to make sense of (or more accurately, to handwave away the fact that it doesn't make sense). This ties in with
8.' $\langle x' | x\rangle$, the braket between different positions $x,x'\in R$, which evaluates to $\delta(x-x')$. Of course, this then means that
8. $\langle x | x\rangle$, with both positions equal, is not actually defined.
If this looks like physicists not caring about rigour in any way, it's because it mostly is. I should stress, though, that it is possible to give a rigorous foundation to these states, through a formalism known as rigged Hilbert spaces, where you essentially split $\mathscr H$ and $\mathscr H^*$ into different "layers". On balance, though, this requires more functional analysis than most physicists really learn, and it's not required to successfully operate on these objects.
Having done, that, we now come to some of the places where you've gone down some very strange roads:
3. $\langle x| O\rangle$ does not mean anything. Neither does "operator on an eigenvalue of $O$ that produces the corresponding eigenfunction under a position basis".
4. $|O\rangle$ is not a thing. You never put operators inside a ket (and certainly not on their own).
Operators always act on the outside of the ket. So, say you have an operator $O:\mathscr H\to\mathscr H$, which in mathematician's notation would take a vector $f\in \mathscr H$ and give you another $O(f)\in \mathscr H$. In Dirac notation you tend to put a hat on $\hat O$, and you use $\hat O|f\rangle$ to mean $O(f)$.
In particular, this is used for the most fundamental bit of notation:
- $\langle f |\hat O|g\rangle$, which a mathematician would denote $\varphi_f(O(g)) = (f,O(g))$, or alternatively (once you've defined the hermitian conjugate $O^*$ of $O$) $\varphi_{O^*(f)}(g) = (O^*(f),g)$.
This includes as a special case
5. $\langle f |G(\hat x)|f\rangle$. This is sometimes abbreviated as $\langle G(\hat x)\rangle$, but that's a good recipe for confusion. In this case, $G:R \to \mathbb C$ is generally a function, but $G(\hat x)$ is a whole different object: it's an operator, so e.g. $G(\hat x)|f\rangle$ lives in $\mathscr H$, and its action is such that this vector has wavefunction
$$ \langle x| G(\hat x) | f \rangle = G(x) f(x).$$
The general matrix element $\langle g |G(\hat x)|f\rangle$ is then taken to be the inner product of $|g\rangle$ with this vector, i.e. $\int_R g(x)^*G(x) f(x)\mathrm dx$, and similarly in the special case $g=f$.
Finally, this brings us to your final two questions:
10. The statement that "the bra portion of the bra-ket is always a dummy variable" is false. As you have seen, $\langle f|$ is perfectly well defined. (Also, $x$ and $p$ are not "dummy" variables, either, again as you have seen above.)
11. Similarly, the statement that "the ket portion is always a function/operator" is also false. You never put operators inside a ket (you put them to the left), and it's generally OK to put $x$'s in there (though, again, this does require either more work to bolt things down, or a willingness to handwave away the problems).
I hope this is enough to fix the problems in your understanding and get you using Dirac notation correctly. It does take a while to wrap one's head around but once you do it is very useful. Similarly, there's plenty of issues in terms of how we formalize things like position kets like $|x\rangle$, but they're all surmountable and, most importantly, they make much more sense once you've been using Dirac notation correctly and comfortably for a while.
Best Answer
One way you can show the completeness relation without bra-ket notation is just $$\sum_{i} \langle \psi_i , v \rangle \psi_i = v \qquad \forall v\in\mathcal{H},$$ where $\mathcal{H}$ is the Hilbert space in question, and $\langle \cdot,\cdot\rangle$ is the corresponding inner product.
Or you can say that the linear operator $$\begin{cases} T:\mathcal{H} \rightarrow \mathcal{H}\\[7pt] v \mapsto T(v)=\sum_{i} \langle \psi_i , v \rangle \psi_i\end{cases} $$ is the identity operator $T = 1\!\!1$.