Here's how you get from one to the other. Let me take the case of a particle moving in one dimension.
In this case, we assume that there exist vectors $|x\rangle$ which form a "dirac-normalized" basis (the position basis) for the Hilbert space in the sense that their inner products satisfy
$$
\langle x|x'\rangle =\delta(x-x')
$$
Note, as an aside, that these vectors are not normalizable in the standard sense, and therefore they do not strictly speaking belong to the Hilbert space.
Next, for each $|\psi\rangle$ in the Hilbert space, we define the position basis wavefunction $\psi$ corresponding to the state $|\psi\rangle$ as
$$
\psi(x) = \langle x|\psi\rangle
$$
So really, the value $\psi(x)$ of the position basis wavefunction $\psi$ at a point $x$ can simply be thought of as the basis component of $|\psi\rangle$ in the direction of $|x\rangle$ just as in the finite-dimensional case where one can find the component of a vector $|\psi\rangle$ along a basis vector $|e_i\rangle$ simply by taking the inner product $\langle e_i|\psi\rangle$.
The inner product is always between two (ket) vectors. However, let $\mathcal H$ be the space in which they live. This is a complex vector space with a Hermitian inner product. The inner product defines a map $\mathcal H\to\mathcal H^\vee$, where $\mathcal H^\vee$ is the dual of $\mathcal H$, and consists of linear functionals on $\mathcal H$. The map is given by $a\mapsto a^\ast$ which is defined by $a^\ast(b) = \langle a,b\rangle$.
In fact the inner product defines a metric on $\mathcal H$ and the postulates of quantum mechanics state that this state space is in fact complete, making it a Hilbert space, and it can be seen that the element $a^\ast = \langle a,\,\cdot\,\rangle$ is continuous. Now denote by $\mathcal H^\ast$ the topological dual of $\mathcal H$, consisting only of continuous linear functionals. The Riesz representation theorem asserts that $a\mapsto a^\ast$ is an isometric isomorphism between $\mathcal H$ and $\mathcal H^\ast$. You should see the latter as the space of bras, and so we see that indeed every bra vector is the conjugate of a ket vector. Starting from that point, in physics one forgets about the distinction between $\langle a,b\rangle$ and $a^\ast(b)$, which are both denoted $\langle a|b\rangle$. It is even customary for this reason to write $\langle a|$ for $a^\ast$ and $|b\rangle$ for $b$.
ADDED IN EDIT
As commented by ACuriousMind, it can be enlightening to make this concrete for the finite dimensional case. In the finite dimensional case, if we fix a basis, $a$ and $b$ can be written as
$$a = \begin{pmatrix}a_1\\ \vdots \\ a_n\end{pmatrix},\ \ \ b = \begin{pmatrix}b_1\\ \vdots \\ b_n\end{pmatrix}.$$
The inner product is $\langle a,b\rangle = a_1^\ast b_1 + \cdots + a_n^\ast b_n$, where $a_i^\ast$ is the complex conjugate of $a_i$. The linear functional $a^\ast$ is represented by a matrix in this basis, namely
$$a^\ast = \left(a_1^\ast\ \cdots\ a_n^\ast\right).$$
Clearly we have $a^\ast(b) \equiv a^\ast b = \langle a,b\rangle$.
ADDED IN SECOND EDIT
The above argument is a strictly mathematical one, in which the (common) assumption is made that kets are the elements of some Hilbert space and bra's are defined to be elements of its continuous dual. This definition is mathematically convenient but not physicaly unavoidable.
First of all, only kets directly relate to physical reality, bra's are just there for mathematical convience. The convenience, and the whole usefulness of the Dirac formalism disappears when you abandon the duality between bra's and kets.
ZeroTheHero remarked that in the book of Cohen-Tannoudji e.a. the definition of bra's and kets (for a single spinless particle) are such that this duality doesn't strictly hold. The authors do acknowledge the computational desirability of this duality, and propose a formal solution, in which the duality is supposedly restored, but with the understanding that not all elements are physical (only aproximately so), and without going into details. I do think they take great care to indicate possible pain points in the formalism, and probably they do the right thing by not going into distracting detail.
My guess is that their intention was to give a pragmatic approach in which physical motivation and intuition are preferred over mathematical rigor. To be specific, first of all they define the Hilbert space of kets $\mathscr F$ to be a subspace of $L^2$ of "sufficiently regular functions" of which they "shall not try to give a precise, general list of [...] supplementary conditions".
Then they define the space of bra's as the space of all linear functionals on $\mathscr F$, the space that in my earlier notation would have been $\mathscr F^\vee$. I don't think it is ever useful to consider the entire (algebraic) dual in the infinite-dimensional case, since this space is huge, and full of highly pathological elements. They'd better have "defined" it as an otherwise unspecified subspace of "sufficiently regular" linear functionals.
Then they construct a very reasonable bra (namely evaluation at a point), and show that it is not associated to any ket, and finally they note that this can be physically resolved by going to generalized kets to restore the duality, even though one should not attribute a physical meaning to them. Many reasonable bra's correspond to a generalized ket, but this is still not the case for the large majority of the elements of $\mathscr F^\vee$.
Best Answer
1) Whenever one has a topological vector space (TVS) $V$ over some field $\mathbb{F}$, one can construct a dual vector space $V^*$ consisting of continuous linear functionals $f:V\to\mathbb{F}$.
2) Under relative mild conditions on the topology of $V$, it is possible to turn the dual vector space $V^*$ into a TVS. One may iterate the construct of dual vector spaces, so that more generally, one may consider the double-dual vector space $V^{**}$, the triple-dual vector space $V^{***}$, etc.
3) There is a natural/canonical injective linear map $i :V\to V^{**}$. It is defined as
$$i(v)(f):=f(v),\qquad\qquad v\in V, \qquad\qquad f\in V^*. $$
4) If the map $i$ is bijective $V\cong V^{**}$, one says that $V$ is a reflexive TVS.
5) If $V$ is an inner product space (which is a particularly nice example of a TVS), then there is a natural/canonical injective conjugated linear map $j :V\to V^*$. It is defined as
$$j(v)(w):=\langle v, w \rangle ,\qquad\qquad v,w\in V. $$
Here we follow the Dirac convention that the "bracket" $\langle\cdot, \cdot \rangle$ is conjugated linear in the first entry (as opposed to a lot of the math literature).
6) Riesz representation theorem (RRT) shows that $j$ is a bijection if $V$ is a Hilbert space. In other words, a Hilbert space is selfdual $V\cong V^*$. If one identifies $V$ with the set of kets, and $V^*$ with the set of bras, one may interpret RRT as saying that there is a natural/canonical one-to-one correspondence between bras and kets.