Let me first give you a heuristic "reason", why the regulator in the class number formula looks different from the regulator in the Birch and Swinnerton-Dyer conjecture. It is often more convenient (and more canonical) to combine the regulator and the torsion term in the Birch and Swinnerton-Dyer conjecture: one chooses a free subgroup $A$ of the Mordell-Weil group of $E(K)$ with finite index and looks at the quantity $R(A)/[E(K):A]^2$, where $R(A)$ is the absolute value of the determinant of the Néron-Tate height pairing on a basis of $A$. As you can easily check, the square in the denominator insures that this quantity is independent of the choice of $A$. Now, in the class number formula, the torsion term is replaced by the number of roots of unity in $K$ and that term is not squared. So for such a canonical formulation to be possible, it is reasonable to expect that not the regulator of the number field should be defined through a symmetric pairing on the units, but its square. Then you could make the same definition as in the elliptic curves case, and it would be independent of the choice of finite index subgroup.
Now, that we have established this, there are several ways to bring the two situations closer together.
You could do the naïve thing: take the matrix $M=(\log|u_i|_{v_j})$, where $u_i$ runs over a basis of the free part of the units (or more generally $S$-units for any set of places $S$ which includes the Archimedean ones), and where $v_j$ runs over all but one Archimedean place (or more generally all but one place in $S$), v_0, say. The absolute values have to be suitably normalised (see e.g. Tate's book on Stark's conjecture). Now take the symmetric matrix $MM^{tr}$ and define this to be the matrix of a new pairing on the units. In other words, you would define your symmetric pairing as
$$
(u_1,u_2) = \sum_{v\in S\backslash\{v_0\}} \log|u_1|_v\log|u_2|_v.
$$
Then, it's clear that you have a symmetric pairing and that the determinant of that pairing with respect to any basis on the free part of the units is $R(K)^2$. As I mentioned above, the square was expected.
Depending on what you want to do, this pairing might not be the best one to consider. For example if now $F/K$ is a finite extension and you consider the analogous pairing on the $S$-units of $F$ and restrict it to $K$, then it's not clear how to compare it to the pairing on $K$. In the elliptic curves case by contrast, the former is $[F:K]$ times the latter. To fix this, we can make the following definition:
$$
(u_1,u_2) = \sum_{v\in S} \frac{1}{e_vf_v}\log|u_1|_v\log|u_2|_v,
$$
where $e$ and $f$ are the absolute ramification index and residue field degree respectively. With this definition, the compatibility upon restriction to subfields is the same as in the elliptic curves case. On the other hand, the relationship with the actual regulator is slightly less obvious. It is however quite easy to show (and I have done it in http://arxiv.org/abs/0904.2416, Lemma 2.12, in case you are interested in the details) that the determinant of this pairing is given by
$$
\frac{\sum e_vf_v}{\prod e_vf_v}R(K)^2,
$$
with both the sum and the product again ranging over the places in $S$.
So I guess, the moral is that one shouldn't seek analogies between the BSD-formula and the class number formula but rather between the BSD and the square of the class number formula (note that also sha, which is supposed to be the elliptic curve analogue of the class number, has square order whenever it is finite). There is also a corresponding heuristic on the analytic side.
Here is a complete proof: as remarked in the answer by Norondion, we can reduce to
the case when $C\_1 \rightarrow C\_2$ is generically separable, i.e. $k(C\_1)$ is separable
over $k(C\_2)$. Let $A \subset k(C\_1)$ be a finite type $k$-algebra consisting of the regular
functions on some non-empty affine open subset $U$ of $C\_2$ (it doesn't matter which one
you choose), so that $k(C\_2)$ is the fraction field of $A$.
By the primitive element theorem, we may write $k(C\_1) = k(C\_2)[\alpha]$, where
$\alpha$ satisfies some polynomial $f(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1} + \cdots
+ a_1 \alpha + a_0 = 0,$ for some $a_i$ in $K(C\_2)$.
Now the $a_i$ can be written as fractions involving elements of $A$, i.e. each
$a_i = b_i/c_i$ for some $b_i,c_i \in A$ (with $c_i$ non-zero). We may replace
$A$ by $A[c\_0^{-1},\ldots,c\_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at
the zeroes of the $c_i$), and thus assume that in fact the $a_i$ lie in $A$.
The ring $A[\alpha]$ is now integral over $A$, and of course has fraction field equal
to $k(C_2)[\alpha] = k(C_1)$. It need not be that $A[\alpha]$ is integrally closed,
though. We are going to shrink $U$ further so we can be sure of this.
By separability of $k(C_1)$ over $k(C_2)$, we know that the discriminant $\Delta$
of $f$ is non-zero, and so replacing $A$ by $A[\Delta^{-1}]$ (i.e. shrinking $U$
some more) we may assume that $\Delta$ is invertible in $A$ as well.
It's now not hard to prove that $A[\alpha]$ is integrally closed over $A$. Thus
$\text{Spec }A[\alpha]$ is the preimage of $U$ in $C_1$ (in a map of smooth curves,
taking preimages of an affine open precisely corresponds to taking the integral closure
of the corresponding ring).
In other words, restricted to $U \subset C_2$, the map has the form
$\text{Spec }A[\alpha] \rightarrow \text{Spec }A,$ or, what is the same,
$\text{Spec }A[x]/(f(x)) \rightarrow \text{Spec A}$.
Now if you fix a closed point $\mathfrak m \in \text{Spec }A,$ the fibre over this point
is equal to $\text{Spec }(A/\mathfrak m)[x]/(\overline{f}(x)) = k[x]/(\overline{f}(x)),$
where here $\overline{f}$ denotes the reduction of $f$ mod $\mathfrak m$.
(Here is where we use that $k$ is algebraically closed, to deduce that $A/\mathfrak m = k,$
and not some finite extension of $k$.)
Now we arranged for $\Delta$ to be in $A^{\times}$, and so $\bar{\Delta}$ (the reduction
of $\Delta$ mod $\mathfrak m$, or equivalently, the discriminant of $\bar{f}$)
is non-zero, and so $k[x]/(\bar{f}(x))$ is just a product of copies of $k$,
as many as equal to the degree of $f$, which equals the degree of $k(C_1)$ over
$k(C_2)$. Thus $\text{Spec }k[x]/(\bar{f}(x))$ is a union of that many points,
which is what we wanted to show.
Best Answer
The reason is that there are two ways of thinking about "points".
Let $A$ be a ring. Then, define:
Note that if $A$ is a $k$-algebra, then geometric points of Spec $A$ are just morphisms Spec $\overline{k}\rightarrow\text{Spec }A$. Further, if $A$ is a finite type $k$-algebra (for example $k[x,y]/(f)$) then a geometric point of $\text{Spec }k[x,y]/(f)$ is nothing but a pair $(a,b)\in\overline{k}^2$ satisfying the equation $f$, which is precisely how Silverman defines a "point".
Here's an example that will explain everything:
Let $k = \mathbb{Q}$, and let $A = \mathbb{Q}[x]$, $B = \mathbb{Q}[y]$, and suppose $X = \text{Spec }A$ and $Y = \text{Spec }B$. Consider the map $f : Y\rightarrow X$ given by $x\mapsto y^2$. This map is degree 2, and sends $y = a$ in $Y$ to $x = a^2$ in $X$.
First, lets consider the scheme-theoretic/topological point $q\in X$ corresponding to $x = -1$. Thus, $q$ is the prime ideal $(x+1)\subset\mathbb{Q}[x]$. It's not hard to check that the only prime of $B$ lying over $q$ is $p := (x^2+1)$, whose residue field is $\mathbb{Q}(i)$, so $p$ has "inertia degree" 2 over $q$.
Now, lets consider the geometric point $Q : $Spec $\overline{\mathbb{Q}}\rightarrow$ Spec $A$ given by sending $x\mapsto -1$. It's not hard to see that sending $y\mapsto i$ and sending $y\mapsto -i$ define two distinct morphisms $P_1,P_2 : \overline{\mathbb{Q}}\rightarrow$ Spec $B$ (ie, two distinct geometric points) which lie above $Q$.
Thus, while there is only one topological point above $x = -1$, there are two geometric points, each of which makes an appearance as a summand of the equation in Prop 2.6. This makes up for the fact that the inertia degree doesn't appear. In fact, one may define the inertia degree as the number of geometric points lying over $x = -1$ who have the same image $P\in Y$.
Ie, in the situation of our morphism $f : Y\rightarrow X$, the two equivalent ways of writing the formula would be: