I think this does it.
Let $\{h_i\}_{i\in I}$ be a linearly independent subset of elements of infinite order of $H$, and given a linearly independent subset $S$ of elements of infinite order of $G/H$, say indexed by $J$, for each $j\in J$ let $g_j$ be an element of $G$ such that $g_jH$ is an element of $S$, so $S=\{g_jH\}_{j\in J}$. Note that because $g_jH$ is of infinite order in $G/H$, then $g_j$ is of infinite order in $G$. Note also that $\langle g_j\mid j\in J\rangle\cap H$ must be trivial: otherwise, there exist $j_1,\ldots,j_k\in J$, and nonzero integers $a_1,\ldots,a_k$, such that $a_1g_{j_1}+\cdots+a_kg_{j_k}\in H$, which means this combination is trivial in $G/H$, which means that $a_tg_{j_t}$ is trivial in $G/H$, which means that $g_{j_t}$ has finite order in $G/H$, a contradiction.
Let us consider $\{h_i\}_{i\in I}\cup\{g_j\}_{j\in J}$. I claim that this is linearly independent in $G$. Indeed, say we have a linear combination of elements of $H$ and $g_j$ equal to $0$; but because $\langle g_j\mid j\in J\rangle$ intersects $H$ trivially, this combination leads to a linear combination of $h_i$ equal to $0$ and one of the $g_j$ equal to $0$. So these are themselves equal to $0$, and hence are the trivial linear combination.
This shows that $r_0(G)$ is at least $r_0(H)+r_0(G/H)$.
Now let $\{g_k\}_{k\in K}$ be a maximal linearly independent set of elements of infinite order in $G$. It is possible that some of the $g_k$ have finite order in $G/H$, so divide $K$ into a disjoint union, $K = K_1\cup K_2$, where for each $k\in K$, we have that $k\in K_1$ if and only if $g_k$ has finite order in $G/H$.
If $k\in K_1$, then there exists $r\gt 0$ such that $rg_k\in H$; but the element is nontrivial, since $g_k$ has infinite order in $G$. Fix such an $r$ and let $h_k=rg_k$. Note that because $\{g_k\}_{k\in K_1}$ is independent, it follows that $\{h_k\}_{k\in K_1}$ is an independent subset of $H$. Thus, $|K_1|\leq r_0(H)$.
We are now going to do a bit of a shuffle: it's possible for the $g_k$ with $k\in K_2$ to not be linearly independent in $G/H$ (here's the example I had in mind: take $G=\mathbb{Z}\oplus\mathbb{Z}$, $H=\mathbb{Z}\oplus\{0\}$, and $g_1=(1,1)$, $g_2=(0,1)$. Then both elements are of infinite order in $G/H$, but they are not independent).
Let's well-order $K$, since you are okay with the Axiom of Choice. I wonder if the following can be justified without this, but I don't know and this seems straightforward, so here goes.
If $\{g_kH\mid k\in K_2\}$ is linearly independent in $G/H$, we are done. Otherwise, there is a least index $r\in K_2$ such that $\{g_kH\mid k\in K_2,k\leq r\}$ is linearly dependent in $G/H$. Thus there is a nontrivial linear combination $0\neq h_r=\sum_{i\leq r}a_ig_i \in H$, with $a_r\neq 0$. Note that $h_r$ is linearly independent from $\{h_k\mid k\in K_1\}$, because the original family was linearly independent. Thus, if we replace $g_r$ with $h_r$, we obtain a set that remains linearly independent in $G$, and we've moved an index from $K_2$ to $K_1$. The least index in $K$ that witnesses that $\{g_kH\mid k\in K_2\}$ is linearly dependent in $G/H$ is now larger than $r$.
This process allows us to set up a transfinite induction on the well-ordered set $K$ that will replace our original collection with a new collection that is still linearly independent, still contains only elements of infinite order, is still partitioned into $K_1$ and $K_2$, but now $\{g_kH\mid k\in K_2\}$ is linearly independent in $G/H$. Because linear combinations are inherently finite, the limit step is exactly the same as the regular step.
At the end of the transfinite induction, we have a disjoint union $K=K_1\cup K_2$, with $\{h_k\mid k\in K_1\}$ linearly independent in $H$, and $\{g_kH\mid k\in K_2\}$ linearly independent in $G/H$. This shows that $|K_1|\leq r_0(H)$ and $|K_2|\leq r_0(G/H)$, hence $|K|\leq r_0(H)+r_0(G/H)$. Thus, $r_0(G)$ is bounded above by $r_0(H)+r_0(G/H)$, proving the other inequality.
