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 (finite dimensional, although generalizable to infinite dimensions in nice enough settings like QM) vector space $V$ with basis $(v_1, v_2,\ldots,v_n)$, a linear functional $f:V\to\Bbb F$ is uniquely determined by how it acts on the basis vectors $v_i$. That's because any vector $v\in V$ may be written as a linear combination $v = a_1v_1+\cdots a_nv_n$ with $a_i\in \Bbb F$, and $f$ is linear, so
$$
f(v) = f(a_1v_1+\cdots a_nv_n) = a_1f(v_1) + \cdots +a_nf(v_n)
$$
so $f(v)$ is completely determined by the $f(v_i)$, for any vector $v$. Thus any $f$ can be represented by a tuple (list of numbers) $(f_1,\ldots,f_n)\in \Bbb F^n$ where $f_i = f(v_i)$. Clearly, if two functions are different, then their corresponding tuples are also different.
On the other hand, any such tuple $(g_1, \ldots,g_n)$ may be used to define a function $g:V\to \Bbb F$. What's more, any such $g$ is linear.
So now we have a correspondence between the set of linear functionals $V\to \Bbb F$ and tuples in $\Bbb F^n$. These tuples are what becomes row vectors in the case where $V$ is a vector space of column vectors.
Best Answer
The set $V^*$ of all linear maps from a vector space $V$ to its base field $K$ is, itself, a vector space over $K$: this is easy to show, with the operations being $(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda(f(x))$, for all $f, g\in V^*$ and all $\lambda \in K$.
Now, given any $\varphi \in W^*$, and any $f: V \to W$, $\varphi\circ f$ is a linear map from $V$ to $K$ (since it's a composition of linear maps), so is an element of $V^*$ (you have a typo in your question: it is not an element of $V$).
Thus, we define a new function, $f^*$, that takes as its input an element $\varphi$ of $W^*$, and spits out an element of $V^*$, by composing $\varphi$ with $f$. This, it turns out, is linear, and this, too, is not hard to check: if $\varphi,\psi\in W^*$, and $v \in V$, then
\begin{align*}f^*(\varphi+\psi)(v) &= (\varphi + \psi)\circ f(v) \\&= (\varphi+\psi)(f(v)) \\&= \varphi(f(v)) + \psi(f(v)) \\&= \varphi\circ f(v) + \psi \circ f(v) \\&= (f^*\varphi + f^*\psi)(v),\end{align*}
hence $f^*(\varphi + \psi) = f^*\varphi + f^*\psi$. Similarly, for any $\lambda \in K$, we have
\begin{align*}f^*(\lambda\varphi)(v) &= (\lambda\varphi)\circ f(v) \\&= (\lambda\varphi)(f(v))\\&=\lambda(\varphi(f(v))) \\&= \lambda(\varphi \circ f)(v)) = \lambda f^*(\varphi)(v),\end{align*}
hence $f^*(\lambda\varphi) = \lambda f^*(\varphi)$, and $f^*$ is, indeed, linear.