This is the more important case of "even" lattices, where all inner products are integral and all vector norms are even. We follow pages 131-134 in Wolfgang Ebeling, Lattices and Codes, available for sale at LINK
\$45.00 for paperback.
Given an even positive lattice $\Lambda$ with covering radius below $\sqrt 2,$ form the integral Lorentzian lattice $L = \Lambda \oplus U.$ Elements are of the form
$ (\lambda, m,n) $ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by
$$ (\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product
$$ (\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1 \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$
We choose a particular set of roots (elements of norm 2) beginning with any $\lambda \in \Lambda$ by
$$ \tilde{\lambda} = \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$
Let the group $ R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\lambda}$ and by $\pm 1.$
Given any $ l \in L,$ a primitive null vector, we have $l \cdot l = 0,$ and so $l \in l^\perp.$ Furthermore, if
$k \in l^\perp$ as well, then $ (k + l)^2 = k^2.$ As a result, we may form the lattice spanned $\langle l \rangle$ by $l$ itself, then form another lattice with norm, $$ E(l) = l^\perp / \langle l \rangle.$$ It took me a bit of doing to confirm (a calculus exercise along with Cauchy-Schwarz) that $E(l)$ is positive definite, all norms are positive except for the $0$ class.
Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$ s_r (z) = z - ( r \cdot z) r.$$
Also $ s_r^2(z) = z.$
It is an exercise to show that, when $ s_r(z) = y,$ then $E(z)$ and $E(y)$ are isomorphic.
There is rather more than first appears to the question:
Lorentzian characterization of genus
In particular, every even lattice in the same genus as $\Lambda$ occurs as some $E(u),$ where
$u \in L$ is a primitive null vector. Furthermore, taking $w = (0, 0,1),$ we find that
$E(w) = \Lambda.$
So, all we really need to do to prove class number one is show that every primitive null vector can be taken to $w = (\vec{0}, 0,1),$
by a sequence of operations in $R.$
Given some primitive null vector
$$ z = (\xi, a,b),$$
so that $ 2 a b = - \xi^2.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$
In the first case, suppose $|b| < |a|.$ Then $ | 2 a b| = \xi^2 < 2 a^2,$ so in fact
$$ \left( \frac{\xi}{a} \right)^2 < 2.$$
We choose the root $ \tilde{\lambda} = \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and
$$ s_{\tilde{\lambda}} (z) = z - ( \tilde{\lambda} \cdot z) \tilde{\lambda} = (\xi, -b,-a).$$
Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that
$ b = \frac{- \xi^2}{2a}.$ Now we have
$ 2 a^2 \leq | 2 a b| = \xi^2,$ so $ \left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some
nonzero vector $\lambda \in \Lambda$ such that the rational number
$$ \left( \frac{\xi}{a} - \lambda \right)^2 < 2. $$
We form the root from this $\lambda,$ as in
$$ \tilde{\lambda} = \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$
For convenience we write
$$ a' = \frac{a}{2} \left( \frac{\xi}{a} - \lambda \right)^2 $$
From the equation $z \cdot \tilde{\lambda} = a - a',$ we see that $a' \in \mathbb Z.$
Well, we have $$ s_{\tilde{\lambda}} (z) = (\xi - ( a - a') \lambda, a',b'), $$ where
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right).$$
Now, if $a'=0,$ then $ s_{\tilde{\lambda}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to
$w = (0, 0,1),$ with
$E(w) = \Lambda.$
If, instead, $a' \neq 0,$ note that our use of the covering radius condition shows that
$ | a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and
$| a - a'| < |a|.$ From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \lambda)^2,$ so
$b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\lambda^2 \geq 2,$ so
$$ 1 - \frac{\lambda^2}{2} \leq 0, $$
and $ ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$ has the same sign as $b$ and $b'.$
From
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$$
we conclude that $| b'| \leq |b|.$
So, these steps have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w.$ So, actually, the covering radius hypothesis implies that all primitive null vectors are mapped to $w$ by a finite sequence of reflections, so that all the $E(z)$ are in fact isomorphic to $E(w) = \Lambda.$ That is, all lattices in the genus are in fact isomorphic, and the class number is one.
Best Answer
One problem that seems to be implicit in your question is that the term "lattice" is used in many contexts, and has multiple definitions. Among people who work with integral bilinear forms or quadratic forms, the norm on a lattice is defined to take values in integers, but it is definitely not assumed to be positive definite (contrary to the definition you offered at the top).
The most classical arithmetic origins come from Brahmagupta's 7th century work on binary quadratic forms, leading to Gauss's composition law and questions related to Waring's problem about which natural numbers are represented by quadratic forms of a certain type. As others have mentioned, there are applications that arise quite naturally in studying the cohomology (in particular, intersection theory) of manifolds, and in number theory proper, where they arise in the study of modular forms via theta constants and trace forms. If you want to study orthogonal groups or Clifford algebras in a setting that includes both real coefficients and finite fields, it is necessary to consider quadratic forms defined over number rings, and in particular, lattices that take values in the integers. A more recent application is in lattice conformal field theory, where you need the bilinear form on a lattice to be integer-valued to yield a super vector space of states (and it must be even to yield an honest vector space).
In the case of complex elliptic curves, the integrality of the lattice (after suitable rescaling) is equivalent to the elliptic curve having complex multiplication. There are plenty of interesting things to say about how these curves relate to class field theory, but it doesn't have much to do with the curves being defined over $\mathbb{Q}$. Most curves over $\mathbb{Q}$ do not have CM, and most CM curves are not defined over $\mathbb{Q}$. See chapter 2 in Silverman's Advanced Topics.