The bra $\langle b |$ denotes the state of the system that can be observed in an experiment,
It is true that usually how inner products $\langle b | \psi \rangle$ show up is in a context where the leftmost ket is the observable quantity and $\psi$ is the state of the system. But just like any math you can (and probably will) see inner products calculated in which this need not be the case. The important thing to remember here is the following, which is one of the core "postulates" of quantum mechanics: If the particle is in a state $|\psi \rangle$, then the probability of getting some experimental outcome, say an energy of $E$, is equal to $|\langle E | \psi \rangle|^2 $, where $|E\rangle$ is a vector corresponding to that experimental outcome. You can calculate this for specific energies, so the probability of getting outcome $E=0.2$ Joules is $|\langle 0.2 | \psi \rangle|^2$, as long as that is one possible energy that the system is compatible with. If there is a continuum of outcomes rather than a countable set of outcomes, then this gives a probability density rather than a probability. That is the reason that you have experimental outcomes showing up in inner products.
Notice, by the way, that since $\langle f | g \rangle = \overline{\langle g|f \rangle}^*$ for any vectors or functions $f,g$, then $|\langle E | \psi \rangle|^2=|\langle \psi | E \rangle|^2$, so in these probabilities we need not have the "experimental outcome ket" on the left, though we normally do.
and it is described by the complex conjugate of the function $f_{b}^{}$, $f_{b}^{*}$.
Given some ket $|\psi\rangle$, then there is a corresponding bra which we write as $\langle \psi |$ which is basically just the same vector, but in bra form instead of ket form. In that case the functions (which give the components of that vector) are just complex conjugates of each other. Certainly you could take the inner product $\langle \psi | \psi \rangle$. However in general if you have an inner product $\langle \phi | \psi \rangle$, the two functions can be completely different. Most physicists think of this in terms of finite-dimensional vectors, wherein if a ket is
$|\psi \rangle = \begin{bmatrix}\psi_1 \\ \psi_2\end{bmatrix}$
Then its corresponding bra is
$\langle \psi | = \begin{bmatrix}\psi_1^* & \psi_2^*\end{bmatrix}$
Bra is achieved by applying an appropriate operator $\Omega$ to the ket.
This is not true. Generally the bra corresponding to a ket is achieved by applying a conjugate transpose, as above.
An operator is a quantum mechanical equivalent of a classical function, and represents an observable in a dynamical variable, e.g. position varying with time.
In quantum mechanics, there is a correspondence between classical variables (not functions) and operators. Position $x$ has its own operator, momentum $p$ has its own operator, so do energy $E$ and angular momentum $L$ have their own operators. The operators don't vary in time, at least not in your introductory course which will use the "Schrödinger Picture" of quantum mechanics. Later you can change pictures, but this is not needed now.
For example, if the initial state of the system is $|k\rangle$, then we can apply the position operator $\Omega$ to obtain a new state $|b\rangle = \Omega | k \rangle$, which represents the position of the particle.
All correct except for the last part - the new state does not represent the position of the particle. The importance of an observable operator like the position operator $\hat{x}$ is basically just that we can use it to find position eigenstates $|x\rangle$ through the eigenvalue equation:
$$\hat{x}|x\rangle = x |x \rangle$$
The labels on that equation can be confusing because there are so many $x$, I'm sorry. One is an operator, one is a vector, one is just a number. Secondly, the position operator can be useful to find expected values. The average value of the position when the particle is in a state $|\psi \rangle$ is equal to $\langle \psi | x | \psi \rangle$.
We can then compare the state $|b\rangle$ obtained from the calculation with the actual result of the experiment to see if they agree.
No this is not correct. The comparison to experiment comes from the predictions of probabilities, which come from repeating the same experiment with the same initial conditions over and over. How to get those probabilities I answered at the beginning of this answer.
Anything I didn't comment on was correct.
