What is the meaning of this notation in the dual space definition? This is just about the clearest definition of double dual space I have come across:

Let $V$ be a vector space over $\mathbb R$. We know that a
  finite-dimensional vector space $V$ is isomorphic to its dual space
  $V^*$. This isomorphism is not ‘very nice’ because it depends on the
  choice of basis. There is however one example of a ‘nice’ or natural
  isomorphism between two vector spaces. This is an isomorphism between
  a finite-dimensional vector space $V$ and the dual space $V^{**}$ of
  its dual $V^*.$ The space $V^{**}$ is called the double (or second)
  dual of $V$. The elements of $V^{**}$ are linear functionals of linear
  functionals. 
Let $v_0$ be a vector in $V$. For each $f \in V^*$, we write
$$v_0^*(f)=f(v_0)$$
In this formula, $v_0$ is fixed (in $V$) and $f$ is allowed to vary
  (in $V^*$), so that the formula defines a function on $V^*$.

But I still don't get it. In part it might be the notation:
What is $v_0^*$ and how do you verbalize the notation? The quote defines $v_0$ as a vector in the original $V$ vector space. That is clear, although the need for the subindex $0$ is either just annoying or its meaning obscure. But there is never a definition of $v_0^*$. I presume it is something along the lines of the "vector being fed to $V^*$ from $V$" (?), but of course then comes the $(f)$, which is defined as a function... and the house of cards inside $v_0^*(f)=f(v_0)$ seems to collapse again.
Is this thing, $a_0^*(f)\in V^{**}?$
Asking for an example may be too much, but a deciphering of the notation would be a great starting point to start thinking about the concept.


CONCLUSIONS (after the accepted answer, comments and additional
  studying):

I initially asked 

how to "verbalize" $v_0^*(f)=f(v_0)$

meaning how to "unpack" this cluster... OK "Charlie Foxtrot" of super and sub-scripted notation in English. This is what I was thinking of as the ideal answer:
It means: Any given vector in $V$, for example, $v_0 \in V$, can be uniquely mapped to an element in $V^{**}$, corresponding to a functional on $V^*$ (that is a functional that eats functions in the space of $V^*$). This functional on $V^{*}$ can be called $v_0^*$. 
That is to say, our new friend, $v_0^*$, belongs to $V^{**}$, but carries one single star because the star makes reference to what it "eats" $(V^{*})$, not where it lives $(V^{**}).$ The author kindly throws the subscript $0$ in $a_0^*$ just in case there was anyone still following at that point - poetic license like in Edgar Allan Poe's “All that we see or seem is but a dream within a dream.” But I digress. In any event, $a_0^*$ lives in the dual of the dual, and as any dual, it is a linear functional. So a function from a real vector space to $\mathbb R.$
The parenthesis, $(f)$ part $v_0^*(f)$, again is telling us that what goes into the functional $a_0^*(\text{here})$ is the functionals $(f)$ that live in $V^*$, sending $V\rightarrow \mathbb R.$ Redundant to have this $(f)$ when we already indicated the "diet" of $a_0^*(f)$ with the asterisk in the superscript? You bet. This part makes sense, though, much like $f(x)=\log x$ makes sense. What is annoying is not having suppressed the superscript $*$ in $a_0^*$. It could have been notated like $L_v(f)$, you know... $L$ for linear functional, $v\in V,$... very simple, but risking the cabalistic aroma of $a_0^*(f).$
OK. So on to the RHS of the equation, which is more of a definition, $:=f(v_0)$
is only uninspired at first glance, being deeply tricky at heart: you look at it and you may assume it the function of... $f(\color{red}{v_0}),$ but it's much more devilish... Here $v_0$ is fixed, and the actual variable is $\color{red}{f(\cdot)}.$ 
$v_0^*(f)=f(v_0)$ evaluates the functionals in $V^*$ at $v_0$, which remains fixed. So the functions of $V^*$ (dual space) are zipped through by the $a_0^*$ functional in $V^{**}$ one at a time, as it where, while $v_0\in V$ stays the same. 
It is in this regard and evaluation map. So if can leave aside the cryptographic notation in the quoted paragraph, and call the function in $V^{**}$ something less obscure, $L_v(f)=f(v)$, where $f\in V^*$ and $v\in V$, we can look at it as a map $L_v: V^*\rightarrow \mathbb R$. Once $v$ is fixed, (let's bring out now the subscripts) to say, $v_0$, the function $L_{v_0}=f(v_0)$ is the evaluation function.
So how to verbalize this? Clearly ‘vee-nought-star' on $X^*.$ As in the answer.
It might be more straightforward to look at it as a dual pair or bracket notation:

If $V$ is ﬁnite dimensional there is a natural “bracketing map” $\phi
> : V^∗ \times V \rightarrow \mathbb R$ given by $\phi:( f,v) \mapsto
> ⟨f,v⟩.$ The expression $\langle f,v \rangle$ is linear in each
  variable when the other is held ﬁxed. If $f$ is ﬁxed we get a linear
  functional $v \mapsto f(v)$ on $V,$ but if we ﬁx $v$ the map $f
> \mapsto \langle f,v \rangle$ is a linear map from $V^∗ \rightarrow
> \mathbb R,$ and hence is an element $j(v) \in V^{∗∗} =( V^∗)^∗,$ the
  “double dual” of $V.$

This last concept is nicely explained here.
 A: The displayed line is the definition of $v_0^*$. Each vector $v_0$ in $V$ determines a linear functional $v_0^*$ (which I read ‘vee-nought-star’) on $X^*$, i.e., an element $v_0^*$ of $V^{**}$. This $v_0^*$ is therefore a linear function from $V^*$ to $\Bbb R$, and it’s defined by
$$v_0^*(f)=f(v_0)\tag{1}$$
for each $f\in V^*$. That is, its value at the linear functional $f\in V^*$ is simply the value of $f$ at $v_0$.
One does of course have to verify that the function from $V^*$ to $\Bbb R$ defined by $(1)$ actually is linear, but this is quite straightforward.
A: We might define a linear transformation $\theta : V \to V^{**}$ by the equation
$$ \theta(v)(f) = f(v) $$
The notation here is recursive; we are defining the linear transformation $\theta$ by specifying its value at every $v \in V$. In turn, we define the linear functional $\theta(v) \in V^{**}$ by specifying its value at every element $f \in V^*$.
To recap, the types of each subexpression are


*

*$\theta : V \to V^{**}$

*$v \in V$

*$f \in V^*$ (equivalently, $f : V \to \mathbb{R}$)

*$\theta(v) \in V^{**}$ (equivalently, $\theta(v) : V^* \to \mathbb{R}$)

*$\theta(v)(f) \in \mathbb{R}$

*$f(v) \in \mathbb{R}$


Now, some people don't like to use the usual function notation for function-valued functions. Here, the author is indicating the function via decoration — the author's notation $v^*$ means the same thing as my notation $\theta(v)$.
Another notation one might use for this is $v \mapsto (f \mapsto f(v))$. If you plug some value $v_0$ into this linear transformation, you get the linear functional $f \mapsto f(v_0)$.  (and if you then plug some linear functional $f_0$ into that, you get the number $f_0(v_0)$)
For some purposes, the most convenient notation is just $v$ written on the right (rather than on the left like functions "usually" are) — i.e. rather than rigidly interpret $f(v)$ as "the function $f$ evaluated at the value $v$", to have the mental flexibility to view it the other way around "$v$ evaluated at $f$", or even "the 'product' of $f$ and $v$", as needed. In many situations, if you can do this, you really don't need to distinguish between $V$ and $V^{**}$ so there is no problem using the same notation both for an element in $V$ and its corresponding element of $V^{**}$.

The reason for $v_0$, I think, is that the author wants you to think of it as an 'unspecified vector constant' rather than a 'vector-valued variable'. In my opinion there is no good reason to do so, but there may be some aspect of the author's choice of fine detail in mathematical grammar or the author's philosophical or pedagogical opinions that compels him to do so.
A: Let me start with a different example. Consider all maps from a set $X$ to $\mathbb{R}$ and push them together in a set, say $M(X,\mathbb{R})$.
This set can be given a structure of vector space by
$$
f+g\colon x\mapsto f(x)+g(x),
\qquad
\alpha f\colon x\mapsto \alpha f(x)
$$
for $f,g\in M(X,\mathbb{R})$ and $\alpha\in\mathbb{R}$.
If $X$ has more structure, then we may identify some useful subspaces of $M(X,\mathbb{R})$. For instance, if $X$ is a metric space, we could consider the continuous maps, $C(X,\mathbb{R})$; if $X$ is a differentiable manifold, we could consider the infinitely differentiable maps, $C^\infty(X,\mathbb{R})$; if $X$ is a measure space, we could consider the integrable maps $\mathscr{L}^1(X,\mathbb{R})$ or the square integrable maps $\mathscr{L}^2(X,\mathbb{R})$.
If we fix $x_0\in X$, we have an interesting map $e_{x_0}\colon M(X,\mathbb{R})\to\mathbb{R}$, simply defined by
$$
e_{x_0}(f)=f(x_0)
$$
Guess what? This map is linear. So we get a map, the evaluation map,
$$
e\colon X\to\operatorname{L}(M(X,\mathbb{R}),\mathbb{R})
\qquad
x_0\mapsto e_{x_0}
$$
The codomain is the set of all linear maps from $M(X,\mathbb{R})$ to $\mathbb{R}$.
Yes, there is another example of interesting subspace I didn't mention before (on purpose). If $X=V$ is a vector space, then the set of linear maps $V\to\mathbb{R}$ is a subspace of $M(V,\mathbb{R})$. This is usually denoted by $V^*$, the dual space of $V$. So the map $e$ is
$$
e\colon X\to M(X,\mathbb{R})^*
$$
In all examples above, $e_{x_0}$ restricts to a map from the “interesting” subspace to $\mathbb{R}$ and so we can define an evaluation map, say,
$$
e\colon X\to\operatorname{L}(C^\infty(X,\mathbb{R}),\mathbb{R})=
C^\infty(X,\mathbb{R})^*
$$
in the case when $X$ is a differentiable manifold. Or, when $X=V$ is a vector space, an evaluation map
$$
e\colon V\to\operatorname{L}(V^*,\mathbb{R})=(V^*)^*=V^{**}
$$
Yes, in this case the codomain is now the dual of $V^*$. And, guess what? The map $e$ is linear (which is easy to prove).
How does the map work? Exactly in the same way as before: it is a particular case, after all: if $v_0\in V$, $e(v_0)$ is an element of $V^{**}$, that is, a linear map $V^*\to\mathbb{R}$; which one? Like before
$$
e\colon V\to V^{**}
$$
sends $v_0$ to $e_{v_0}$, which is defined by
$$
e_{v_0}(f)=f(v_0)
$$
which is exactly the same as saying $e_{v_0}=v_0^*$ as in your notes (although I'd not use a $^*$ here, but probably something like $\widehat{v_0}$).
Getting rid of the noughts, we have a map
$$
e\colon V\to V^{**},\quad v\mapsto e_v
$$
where $e_v(f)=f(v)$.
This can also be seen as a bilinear map $(v,f)\mapsto f(v)$, from $V\times V^*$ to $\mathbb{R}$, which has some interesting uses as well. But, concentrating on $e$, we see that it is injective (provided we accept that every vector space has a basis) and an isomorphisms precisely in the case when $V$ is finite-dimensional.
