A root datum is given by:
- A subset $R$ of a free abelian group $M$
- A subset $C$ of the dual free abelian group Hom$(M,\mathbf{Z})$
- A bijection between $R$ and $C$
subject to conditions. A root system is given by a
- A vector space V over the real numbers $\mathbf{R}$
- A pairing $V \times V \to \mathbf{R}$
- A subset $R$ of $V$
subject to conditions.
A root datum should determine a pair of root systems, one in $M \otimes \mathbf{R}$ and one in Hom$(M,\mathbf{Z}) \otimes \mathbf{R}$. But what is the scalar product on either of these real vector spaces?
The conditions defining a root datum $(M,R,M',C)$ (in your notation, $M'=\mathrm{Hom}(M,\mathbb{Z})$, but I prefer to take a perfect pairing $\left<\cdot,\cdot\right> : M'\times M\to\mathbb{Z}$ between the free abelian groups $M'$ and $M$, which is equivalent to an identification of $M'$ with the dual of $M$) immediately give you that $R$ is a (classical) root system in $M_\mathbb{R}:=M\otimes_\mathbb{Z}\mathbb{R}$, and $C$ is a root system (dual to $R$) in $M'_\mathbb{R}:=M'\otimes_\mathbb{Z}\mathbb{R}$. Now, since the group $M$ is assumed to be free, the perfect pairing $\left<\cdot,\cdot\right>$ extends to $\left<\cdot,\cdot\right> : M'_\mathbb{R}\times M_\mathbb{R} \to\mathbb{R}$ and the bilinear form
$$(x,y):=\sum_{\alpha\in R} \left<\alpha^\vee,x\right>\left<\alpha^\vee,y\right>$$
(where of course $\alpha^\vee\in C$ is the image of $\alpha\in R$ under the bijection $R\to C$) certainly defines an inner product on $M_\mathbb{R}$. Moreover, this inner product also makes sense if you just had a Euclidean root system from the beginning, and coincides with the given Euclidean inner product.