Say I have two smooth one-cycles $Z, Z'$ on a 3-manifold $M$, i.e. just formal linear combinations of (let us say disjoint) smooth circles. Assume they're homologous.
Is there a map to a surface $\pi: M \to S$ and homologous zero-cycles $P, P'$ on the surface such that $\pi$ is a submersion over $P, P'$ and one has $\pi^{-1}(P) = Z$ and $\pi^{-1}(P') = Z'$?
I suspect the answer to this question is no in general.
Suppose $Z$ has two components $Z_1 \cup Z_2$ with different multiplicities (as a formal linear combination of knots) and $Z'$ has one component. Then $P$ must have two points $P=\{p_1,p_2\}$ (ignoring multiplicity), and $P'$ one $P'=\{p'\}$. Then $\pi_1(M-(Z\cup Z'))$ surjects a free group of rank at least two (the meridians of $Z_i$ go to loops about $p_i$, which must generate a subgroup of $\pi_1(S-(P\cup P'))$ of rank at least two).
The corank of a group is the largest rank of a free group that it surjects. It is known that there are 3-manifold groups with arbitrarily large $b_1$ and corank 1. What is needed to get a counterexample to your question is a manifold with 3 torus boundary components which has fundamental group of corank 1, and satisfying an appropriate homological condition. I feel like a construction of the sort that Shelly Harvey carries out ought to have the right properties.
Addendum: I looked at a paper of Leininger-Reid, which gives an example of a compact 3-manifold $M$ with hyperbolic interior, 4 boundary components, and cut number 1. From this, one can get an example which answers your question in the negative. By half-lives, half-dies, there is an oriented surface $\Sigma\subset M$ which is homologically non-trivial in $H_2(M, \partial M)$, and intersects each boundary component in consistently oriented parallel non-trivial curves (in particular, non-empty). Now, perform Dehn fillings on each boundary component so that the intersection number with $\partial \Sigma$ are different to get a manifold $N$. Partition the cores of the Dehn fillings into $Z$ and $Z'$, with multiplicity determined by the intersection number of $\partial \Sigma$ with the meridian of each Dehn filling, and orienting the components of $Z'$ opposite to the induced orientation of $\partial \Sigma$. This gives the desired pair of cycles which are homologous and which cannot be induced by pullback from 0-cycles on a surface because the link complement has cut number 1.
