Suppose that $\{V_{i}\}$ and $\{W_{i}\}$ are directed systems of vector spaces, indexed over $\mathbb{R}$, with all the maps $i_{rr'}: V_{r} \to V_{r'}$ and $j_{rr'}: W_{r} \to W_{r'}$ injective. Thus, all of the vector spaces $V_{i}$ embed in the colimit $\mathbb{V}$ of the their directed system, and similarly with the vector spaces $W_{i}$ and $\mathbb{W}$. Next, suppose that there are morphisms between these directed systems, maps $f_{r}: V_{r} \to W_{r}$ making all the necessary squares commute. Suppose further that each $f_{r}$ is surjective, so we have a surjective map $f: \mathbb{V} \to \mathbb{W}$.
The goal: The construction of a map $g: \mathbb{W} \to \mathbb{V}$ which restricts to each $W_{r}$ as a map $g: W_{r} \to V_{r}$ which is a right inverse for $f_r$. I believe that the injectivity of the maps $j_{rr'}$ should make it possible to do this in a consistent way. However, what confuses me is how to deal with the fact that our indexing set is $\mathbb{R}$, a set on which it is hard to do ordered induction.
Now, if $\mathbb{V}$ and $\mathbb{W}$ have finite rank, this becomes a finite induction task and quite easy. To be honest, that is all I need for the application at hand. However, I am wondering if such a map $g$ can be produced in greater generality. Naturally, AC will have to make an appearance, but I'm not sure if I need any other set-theoretic assumptions.
This is not possible in general, and it has nothing to do with set theory - it comes down to the fact that your diagram is not well-founded.
Let's fix a field $k$ to work over. For all $r\in \mathbb{R}$, let $V_r = \bigoplus_{s\leq r} k$, and let $W_r = k$. For all $r<r'$, let $i_{rr'}\colon V_r\to V_{r'}$ be the obvious inclusion map, and let $j_{rr'}\colon W_r\to W_{r'}$ be $\text{id}_k$. For all $r$, let $f_r\colon V_r\to W_r$ be the map $\bigoplus_{s\leq r} k\to k$ induced by $\text{id}_k$ on each component.
Computing the colimits, we have $V = \bigoplus_{r\in \mathbb{R}} k$, and $W = k$. A map $g\colon W\to V$ is determined by $g(1)$, which has only finitely many nonzero components in $V$. Letting $r$ be the least of these, for any $s<r$, the restriction of $g$ to $W_s$ fails to factor through $V_s$.
Note that the same example works if you replace $(\mathbb{R},\leq)$ with any partial order with no minimal element, e.g. $(\mathbb{Z},\leq)$.