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 $\langle \phi | v \rangle = \langle \phi , v \rangle.$
Given a vector $v$ and a dual vector $f$, you can produce a scalar $f(v)$. This can be viewed as a map $v \mapsto f(v)$ or $f \mapsto f(v)$.
So a vector determines a (linear) map from the space of dual vectors to scalars (i.e. a $(1,0)$-tensor since we have 1 dual vector input and no vector inputs). Likewise a dual vector determines a (linear) map from the space of vectors to scalars (i.e. a $(0,1)$-tensor since we have no dual vector inputs and 1 vector input).
Best Answer
One big difference between GR and QM is that the vector spaces that represent spacetime in GR are real vector spaces, whereas in QM the vector spaces in which quantum states live are complex vectors spaces. So analogies between one field and the other may require some translation.
There is no natural map between a vector space and its co-vector space of linear functionals a.k.a. one-forms - but we are free to choose a map that makes sense in the context of the physical systems that we are studying. In GR the we often choose to use the metric tensor as a map, and the metric tensor (like all tensors) is linear in both its dimensions. In QM we use the complex scalar product which is not bilinear, but is sesquilinear instead.