I assume two things about Dirac’s$^1$ algebraic scheme involving bras and kets.
1: it is a way of thinking and doing the algebra/analysis, that will reproduce the expressions one would obtain using the definitions of ‘Inner Products’ , and ‘Adjoint Operators’, one would find in, Lipschutz and Lipson$^2$, see pgs. 239, and 377, respectively.
2: has one or more advantages when used for quantum mechanical analysis.
In general, the bras corresponding to a given vector space of Dirac kets, may be set up from a knowledge of the usual inner product one would define on such a vector space, see Lipschutz and Lipson$^2$, pg 239.
Example 1
If we had a vector space $V$, of Dirac kets $ |f\rangle=f$ given by
\begin{equation*}
V=\left\{
f|f:\mathbb{R}\to\mathbb{C}, x\mapsto f(x), \int_{- \infty}^\infty f^\ast(x)~f(x)~dx~<\infty
~\right\} \tag{1}\end{equation*}
we would define the usual inner product on $V$, for use in quantum mechanical analysis, $\langle g,f\rangle$, by
\begin{equation*}
\langle g,f\rangle=\int_{- \infty}^\infty g(x)~f^\ast(x)~dx \tag{2}
\end{equation*}
With the use of Dirac’s algebraic scheme in mind we define a Diracian scalar product, $\langle f|g\rangle$, given by
\begin{equation*}
\langle f|g\rangle=\int_{- \infty}^\infty f^\ast(x)~g(x)~dx \tag{3}
\end{equation*}
If we did this, then everytime we had an inner product, working with the “mathematical” notation of Reference 2, we would have
\begin{equation*}
\langle g,f\rangle=\langle f|g\rangle
\end{equation*}
which has us in good shape, for using Dirac's algebraic scheme to reproduce results we would expect when using the “mathematical” notation of Lipschutz and Lipson$^2$.
With $\mathbf(3)$ in mind, it seems OK to set $\langle f|=f^\ast$ when we have $|f\rangle=f$.
Now, I guess, thinking Dirac’s way, we have to think of the functions $f^\ast$ and $f$ ( when $f^\ast$ is a bra and $f$ is a ket ) as being fundamentally different mathematical objects (as being completely different types of vector, which cannot be added together ).
I know they do not look different, they are both just functions of $x$.
Bras and kets are not the only mathematical things that appear in his scheme, the scheme also features operators. The operators are allowed to operate to “the left” as well as in the usual manner, to the right, and when doing so, are to be considered the same operator. The effect of an operator, operating to the left is defined by $(\mathbf{3})$ of pg. 25 of Reference 1. This can lead to a given operator, having different effects on functions, the kets and the bras, depending upon whether it operates to the left, or to the right. This can be made sense of, by considering bras and kets as being fundamentally different types of vector.
We do not consider a bra $\langle g|=g^\ast$ to be the complex conjugate of a ket $|g\rangle=g$, but consider it to be the conjugate imaginary of $|g\rangle=g$.
As I assumed in the question: If we have a ket $|f\rangle$ such that $|f\rangle|_x=f(x)$ then the corresponding bra would be $ \langle f|$ such that $\langle f|~|_x=f^\ast(x)$. This is consistent with ($\mathbf{3}$).
Or put another way: If we had a ket $|f\rangle=f$ then the corresponding bra would be $ \langle f|=f^\ast$.
Example 2
If we start with a space $W$, of spin kets which are two component complex spinors,
\begin{equation*}
W=
\left\{
~ \binom {c_1} {c_2} : ~c_1,c_2\in \mathbb{C}~,~
(c_1^\ast,c_2^\ast) \binom {c_1} {c_2} <\infty
\right\}
\end{equation*}
where we have, using ket notation
\begin{equation*}
\left| \binom {c_1 }{c_2 } \right \rangle= \binom { c_1 }{c_2 }
\end{equation*}
then we would define an inner product, the Lipschutz and Lipson$^2$ way by
\begin{equation*}
\left \langle \binom {c_3 }{c_4 } , \binom {c_1 }{c_2 } \right \rangle= (c_1^\ast,c_2^\ast) \binom {c_3} {c_4}
\end{equation*}
With the use of Dirac’s algebraic scheme in mind, we would define a Diracian scalar product by
\begin{equation*}
\left \langle \binom {c_1 }{c_2 } | \binom {c_3 }{c_4 } \right \rangle=
(c_1^\ast,c_2^\ast) \binom {c_3} {c_4}
\end{equation*}
This definition ensures, that we always have, for any $c_1, c_2, c_3, c_4 \in \mathbb{C}$
\begin{equation*}
\left \langle \binom {c_3 }{c_4 } , \binom {c_1 }{c_2 } \right \rangle=
\left \langle \binom {c_1 }{c_2 } | \binom {c_3 }{c_4 } \right \rangle
\end{equation*}
Guided by the Diracian scalar product, we would take the bra that corresponds to the ket $\left| \binom {c_1 }{c_2 } \right\rangle$ to be
\begin{equation*}
\left\langle \binom {c_1 }{c_2 } \right| = (c_1^\ast,c_2^\ast)
\end{equation*}
Note in this example, how a bra actually looks to be a different type of object to a ket ( row vector as compared to column vector ).
As I assumed in the question:
If we had a spin ket
\begin{equation*}
\left| \binom {c_1} {c_2}\right\rangle = ~ \binom {c_1} {c_2}
\end{equation*}
then the corresponding bra would be
\begin{equation*}
\left\langle\binom{c_1}{c_2}~\right|= ~ ( c_1^*~~~c_2^* )
\end{equation*}
where $c_1,c_2\in \mathbb{C}$.
References.
1, P.A.M. Dirac, The Principles Of Quantum Mechanics 4th Ed., Clarendon Press, Oxford, (1958). Dirac starts to explain his algebraic scheme on pg. 18, in the section entitled ‘6. Bra and ket vectors’
2, Seymour Lipschutz, PhD, Marc Lipson, PhD, SCHAUM’s outlines, Linear Algebra, Fourth Edition, McGraw Hill (2009).
