Let's say I have eigenstates $|x\rangle$ associated with measurement of position. I know that the eigenstates corresponding to their respective eigenvalues form a basis, let's call it $A$. Now let's say I want to expand a generic state vector $|ϕ\rangle$ in that basis formed by eigenstates. Now, what does it mean when physicists say that:
"$ϕ(x)$ is the coordinate of the state $|ϕ\rangle$ in the basis $A$?"
Any explanation or reference to similar kind of questions would be helpful.
If you think of kets $\vert\psi\rangle$ as vectors, i.e. elements of a vector space in the sense that you can take linear combination of them, and you think of these vector as column vectors, then you can think of $\langle\psi\vert$ as a row vector so that $\langle\phi\vert\psi\rangle$ is a scalar product since it gives a number.
If we imagine that there were discretely many $x$'s labelled by $i$, i.e. if our positions came as $\{x_{i-1}, x_i,x_{i+1}\ldots\}$, you could associate the number $\psi(x_i)$ (i.e. the function $\psi$ evaluated at $x_i$) as the component of $\vert \psi\rangle$ on the $i$'th basis vector $\vert x_i\rangle$, with
$\langle x_i\vert\psi\rangle=\psi(x_i)$. In this way, $\vert \psi\rangle$ and $\vert x_i\rangle$ are huge column vectors with components
$$
\vert\psi\rangle = \left(\begin{array}{c}
\vdots \\
\psi(x_{i-1})\\
\psi(x_i)\\
\psi(x_{i-1})\\
\vdots
\end{array}\right)\, ,\qquad
\vert x_i\rangle = \left(\begin{array}{c}
\vdots \\
0\\
1\\
0
\\
\vdots
\end{array}\right)
\begin{array}{c}
\\
\leftarrow\hbox{position $i$}\\
\\
\end{array}\, ,\qquad
\langle x_i\vert\psi\rangle = \psi(x_i)
$$
Of course the position is really continuous, so the index $i$ is not needed but it's still convenient to this of $\psi(x)=\langle x\vert\psi\rangle$ as the value of the vector $\vert\psi\rangle$ on the basis vector $\vert x\rangle$.
I think that what you are asking is not completely a mathematical problem but a physical trick...
Since I think that you got stuck around eq. (19) of the document you uploaded, I'll try to explain what's going on there.
The aim is to expand our state $|\psi \rangle$ in terms of our basis vector $|x\rangle$.
We can rewrite the state $|\psi\rangle$ in the following way
$$|\psi\rangle = \hat{1}|\psi\rangle $$ At this point you only have only multiplied your state by an identity matrix. Then using (18) that is nothing but a vector identity, you can rewrite the product as
$$\sum_k |x\rangle\langle x||\psi\rangle $$
but since in general, a Hilbert space has infinite dimension, we are allowed to consider the integral instead of the sum and then we have
$$|\psi\rangle = \int |x\rangle\langle x||\psi\rangle dx $$
and defining (here's the trick!) $ \psi(x) = \langle x|\psi \rangle$ we finally get the equivalence $$|\psi\rangle = \int \psi(x)|x\rangle dx $$
This $\psi(x) $ is nothing but our familiar wavefunction. In the present language, $\psi(x) $ are the coordinates of the our state $|\psi\rangle$ in the $|x\rangle$ basis, since you're only projecting your function on the element of the basis!
