We have a smooth space-time curve defined by $f:C{\mapsto}M$, where $M$ is a typical curved space-time manifold.
${\eta}^{(4)}$ is the volume 4-form defined on $M$ and ${\varepsilon}^{(1)}$ is the (non-degenerate) volume 1-form induced on $C$. As usual, both forms are defined by the space-time metric.
Making reasonable assumptions about the curve, how does one find its Poincaré dual, denoted by the closed 3-form ${\beta}$.
You should read more carefully the page before 203 (namely 202). $J$ in his example (or $\beta$ in yours) is not a differential 3-form per se. It is a differential 3-form in the distributional sense. You can think of it as
$$ \delta_{f(C)} \iota_{f'} \omega $$
by which I mean a Dirac delta function supported on the curve $f(C)$ times the three form obtained by contracting the velocity vector of $f$ against the space-time volume form.
More generally, the language to think about is that of currents in the sense of de Rham which is used, among other things, in geometric measure theory. The basic idea is to extend distribution theory for functions to a theory for differential forms. So on a $4$-manifold $M$ we can consider, for example, the set $\mathcal{D}_1$ of all continuous linear functionals mapping smooth one forms of compact support to the real numbers. There are two general categories that should be familiar:
The space $\mathcal{D}_1$ is much larger, since it is the dual of a topological vector space, it is in particular a linear space, and we need to start taking linear combinations of objects listed above. The sum of a three form and a 1 dimensional submanifold isn't really either.
But what about the notation $J\wedge A$? If it is not a differential form, why do we take a wedge product?
This is really just an abuse of notation inherited in part from distribution theory. Recall that frequently we write Schwartz distributions as if they were functions (for example, recall that we think about the Dirac delta as a "generalized function" and write $\int_M \delta f \mathrm{d}x$ for the duality pairing between the distribution $\delta$ and the function $f$). Drawing an analogy with the three forms case where the wedge product actually makes sense, sometimes people write $$ \int_M J\wedge \kappa $$ for the duality pairing of evaluating the continuous linear functional $J: \Omega^1_c(M)\to \mathbb{R}$ on the element $\kappa \in \Omega^1_c(M)$.
Now then, what does this have to do with Poincare duality?
Suppose now that $\kappa$ is a closed one-form, and $S$ is a closed curve in $M$. From Stokes theorem the integral $\int_S \kappa$ only depends on the homology class of $S$ and the cohomology class of $\kappa$. And by Poincare duality, there exists actually a canonical isomorphism between homology classes $H_1$ (of closed curves) with the cohomology classes $H^3$ (of closed three forms).
In other words, for every closed curve $S$ in $M$ there exists a (possibly nonunique, up to addition of an exact form) three form $\eta$ such that $$ \int_S \kappa = \int_M \eta\wedge \kappa $$ holds provided that $\kappa$ is closed (that is, $\mathrm{d}\kappa = 0$). Note that for non-closed forms this equality may be invalid. So the above equality is not saying that the images in $\mathcal{D}_1$ of $S$ and $\eta$ are the same thing!
On page 203 when it is mentioned that the "Poincare dual of a world line" is considered, the authors really meant that they are considering the current $J\in\mathcal{D}_1$ (the latter set they seemed to have named $\Omega^{|-1|}$) representing the image of the curve $f(C)$ (which is a smooth one dimensional submanifold). In equation (4.14) of the book the third term on the right $$ \int_{T_1}^{T_2} x^* A $$ where $A$ is a two form and $x: \mathbb{R} \to M$ is a smooth map is precisely the correct mathematical way to write the duality pairing of this $J$ against the one form $A$.
(I dislike this use of the terminology "Poincare duality", but it appears that it is endemic in the string theory and high energy physics community.)