I was trying to read the above section from the book, I am lost. I am confused with notation and terminology.
I understand till here "..... is a group of type $(2,2)$".
Now I am confused with notation. I read on wikipedia that $G_{m}$ denotes the multiplicative group of the field. I am confused as $H^{*}$ is also the multiplicative group of standard field of quaternion. Why use two notation for multiplicative group of standard field of quaternions.
What is this map, denoted by $N$. He says It is the reduced norm.
Edits (Following the comments): 
Here are my best guesses:
1) I think the phrase a group "of type $(2,2)$" just means that the (finitely generated abelian) group's invariant factors are $2,2$, i.e., $$ D = \frac{Q_8}{\{\pm 1\}} \cong \frac{\mathbb{Z}}{2\mathbb{Z}} \times \frac{\mathbb{Z}}{2\mathbb{Z}} \, . $$ Darmon/Serre uses this notation on p. xiv of the Intro, p. 13, and again on p. 16, and it seems consistent with this interpretation.
2) The reduced norm of a quaternion $\alpha = t + xi + yj + zk$ is \begin{align*} N(\alpha) &= (t + xi + yj + zk)\overline{(t + xi + yj + zk)} = (t + xi + yj + zk)(t - xi - yj - zk)\\ &= t^2 + x^2 + y^2 + z^2 \end{align*} where $\overline{{}\cdot{}}$ denotes conjugation, similar to that for complex numbers. (See here for the definition of reduced norm for a general quaternion algebra.)
3) I think $\mathbb{G}_m$ means $\mathbb{G}_m(\mathbb{Q}) = \mathbb{Q}^\times$, that is, the multiplicative group of $\mathbb{Q}$, not $\mathbb{H}$. One can show (again, just like for complex numbers) that a quaternion is a unit iff its norm is nonzero. So the first component of the map $(N, \phi): \mathbb{H}^\times \to \mathbb{G}_m \times \text{SO}_3$ is just the norm map, which takes \begin{align*} N: \mathbb{H}^\times &\to \mathbb{G}_m = \mathbb{Q}^\times\\ t + xi + yj + zk &\mapsto t^2 + x^2 + y^2 + z^2 \, . \end{align*}
(For those curious, the notes referenced in the OP can be found here.)