= Solution
The additive <Abelian group> $G$ is divisible because it is a real <vector space>. Each convex subgroup $C_n$ is divisible too: for $c\ge0$ in $C_n$ and a positive integer $m$, its unique additive quotient $c/m$ satisfies $0\le c/m\le c$, hence belongs to $C_n$. Negative elements follow by taking inverses. Thus all $C_n$ are rational <vector subspaces>, independently of whether the given real scalar multiplication respects the order.
Choose an ordered-group isomorphism $\theta_n:C_{n+1}/C_n\to\mathbb R$. It is rational-linear, since additive homomorphisms between divisible torsion-free groups preserve division by integers. Extend a rational <basis> of $C_n$ to one of $C_{n+1}$; equivalently choose a rational-linear section $s_n:\mathbb R\to C_{n+1}$ of $\theta_n$ composed with the quotient map. Then
$$
C_{n+1}=C_n\oplus s_n(\mathbb R).
$$
Since $C_1=0$ and the union is $G$, each $g\in G$ has a unique expression $g=\sum_{j=1}^N s_j(r_j)$ with only finitely many nonzero coefficients. Define $\Phi(g)=(r_1,r_2,\ldots)$. This is an injective additive <group homomorphism>. If $r_N\ne0$ is the largest active coefficient, the image of $g$ in the ordered quotient $C_{N+1}/C_N$ has sign equal to that of $r_N$. Convexity ensures that this is also the sign of $g$.
The maximum-support convention for the <Hahn group> $V(\mathbb Z_{>0},\mathbb R)$ uses precisely finite supports and the sign of the largest active coefficient. Hence $\Phi$ preserves and reflects the <total order>, and therefore preserves <join> and <meet>. This proves the <finite-support Hahn embedding along a countable convex chain>. In fact every finite coefficient sequence is obtained, so $\Phi$ is an ordered-group isomorphism in this convention; the requested embedding follows.
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