First, there's something called the Riesz Representation Theorem. To understand it, start by fixing a vector $v \in H$. We can now use the dot product to define a continuous linear functional $L_v: H \rightarrow F$ by $$L_v(w) = \langle w,v \rangle.$$ What the Riesz Representation Theorem says is that every continuous linear functional $\phi:H \rightarrow F$ arises in this way! That is, given any element $\phi$ of the dual space $H^\ast$, there is some $v \in H$ so that $\phi(w) = \langle w, v \rangle$. So this is why we can sort of think as linear functionals as elements of the Hilbert space, and vice versa. In more mathematical terms, we say that the dual of $H$ is isomorphic to $H$.
Second, a note on dual spaces: I believe that typically the space $H^\ast$ is defined to be the set of continuous linear functionals. This distinction is important, as there are different types of duals. So far, I've been talking about the topological dual. However, there is an algebraic dual, which I have seen denoted $H^\star$, which is just all linear functionals $H \rightarrow F$, no continuity assumed. The Reisz Representation Theorem concerns only the topological dual. (The duals are actually the same for finite dimensional Hilbert spaces, but I don't believe the Hilbert spaces encountered in QM are.)
Third, adjoints: Given any Hilbert spaces $H, K$, and a continuous linear functional $A: H \rightarrow K$, there is a continuous linear map called the adjoint $A^\ast:K \rightarrow H$ (note that it goes the other way) that is defined by the equation $\langle Av, w \rangle = \langle v, A^\ast w \rangle$. You typically only see the case $K = H$. So no, the adjoint of an operator $A: H \rightarrow H$ is not an element of $H^\ast$, since the members of $H^\ast$ are continuous linear functionals from $H$ into $F$, and $H^\ast$ goes from $H$ into $H$.
Unfortunately, I don't know much about QM, so this last bit is just speculating on how I think the notation works. If you consider kets $|\phi \rangle \in H$ to be an element of the Hilbert space, then there is a continuous linear functional that I suppose you could call $\langle \phi |$ defined by $\langle \phi | v \rangle = \langle v, \phi \rangle.$ And conversely, given a bra $\langle \phi |$, by the Reisz Representation Theorem, there is a bra $| \phi \rangle$ so that $\langle \phi | v \rangle = \langle \phi , v \rangle.$
Just take the $L$ to be the identity operator on a finite dimensional Hilbert space (inner product space). And use the orthonormal basis basis .
If your eigen-space has dimension greater than $1$ then it is always possible. Just take the orthonormal basis of the eigen-space.
Best Answer
The underlying reason is that for a Hilbert space $H$, its dual is again $H$. What this means is that any bounded linear functional $H\to\mathbb C$ is given by the inner product against some fixed vector. So, any linear functional is given by choosing some $v\in H$ and then doing $u\longmapsto \langle v|u\rangle$ (this is known to mathematicians as the Riesz Representation Theorem).
That is, the "kets" are the elements of the space $H$, and the "bras" are the elements of the dual. Now, the adjoint $A^\dagger$ is defined precisely as the operator (on the dual $H^*$, but for us it is again $H$) such that $$\tag1\langle A^\dagger v|u\rangle=\langle v|Au\rangle.$$ When $A$ is selfadjoint, you have $$ \langle v|Au\rangle=\langle Av|u\rangle=\overline{\langle u|Av\rangle}. $$ As a mathematician, it is hard for me to grasp any advantage from talking about bras and kets. It would make a lot more sense to write $v^*u$ instead of $\langle v|u\rangle$. Here, for a vector $v$, the vector $v^*$ is the conjugate transpose, and then $v^*u$ is precisely the matrix product. When there is an operator in the mix, $$ v^*(Au)=v^*Au=(A^*v)^*u, $$ which is the same as $(1)$ but doesn't require any convention other than the mathematical properties of taking adjoints (namely, "conjugate and tranpose").
If you really care about the math behind, given an operator $T:X\to Y$, one can always defined an adjoint $T^*:Y^*\to X^*$ (where $X^*$ is the dual, i.e. the space of bounded linear functionals on $X$) by $$ (T^*g)(x)=g(Tx). $$ Because, as already mentioned, for a Hilbert $H$ the dual is again $H$, we can see the adjoint again as an operator on $H$.