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I am completely new in 3-manifolds but I have to understand their equivalent of intersection pairing in 4-manifolds.

Consider a 3-manifold $M$.We define the linking pairing on the torsion part of $H_1(M,\mathbb{Z})$ as following $$ lk: \text{Tor} H_1(M) \times \text{Tor} H_1(M) \to \mathbb{Q/Z} $$ $$ [a] \times [b] \to \#(a \cap b')/n$$ where $b'$ is a 2-cycle such that its boundary satisfies $\partial b' = nb$.

From the right hand side I see that we are calculating the number of intersection points between a 1-cycle $a$ and a 2-cycle $b'$ in some sense.

The questions I have are the following:

  1. This is probably the most naive question, but why is this defined in the torsion part of the homology? Why does even the homology "need" to have torsion part?

  2. Why can we not define directly the intersection of a 1-cycle and a 2-cycle and have have to go through this weird definition using the boundary?

    If you can direct me to any reference it would be great since a quick search on Google did not yield any fruitful results.

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Recall that for links in $S^3$ the definition of linking number (usually given via planar projections) can be restated as follows.

$\textbf{Definition 1.}$ Suppose that $K_1$ and $K_2$ are two knots in $S^3$. Since $H_1(S^3 - K_2, \mathbb{Z})= \mathbb{Z}$ generated by a meridian $\mu$ of $K_2$, we have that $[K_1]=m \cdot [\mu]$ for some $m \in \mathbb{Z}$. Define $lk(K_1, K_2)=m$.

$\textbf{Definition 2.}$ Suppose that $K_1$ and $K_2$ are two knots in $S^3$. Pick a surface $F \subset S^3$ such that $\partial F = K_2$ and define $lk(K_1, K_2)=\# (K_1 \cap F)$.

Let us concentrate on the second one. What is crucial in this definition is that $[K_2]=0$ in $H_1(S^3, \mathbb{Z})$. In fact this condition guarantees the existence of a surface $F$ bounding $K_2$. From this perspective it is clear how to extend the definition of linking number to the case of knots in a integral homology sphere $Y$ (meaning that $H_1(Y, \mathbb{Z})=0$).

Let us suppose now that $Y$ is a rational homology sphere ($H_1(Y, \mathbb{Q})=0$). In this case the existence of a surface $F \subset Y$ bounding $K_2$ is not guaranteed since it can happen that $[K_2]\not=0$ in $H_1(Y, \mathbb{Q})$. On the other hand since $Y$ is a rational homology sphere we can find $n \in \mathbb{Z}$ such that $n[K_2]=0$, meaning that there exists a surface $F \subset Y$ (with $n$ boundary components) bounding $n$ parallel copies of $K_2$. Define $$ lk_\mathbb{Q}(K_1, K_2)= \frac{ \#(F \cap K_2)}{n}\ .$$

From here it is clear how to define the linking number of two knots a in rational homology sphere, or more in general of two rationally null-homologous knots in a random three-manifold. The funny thing is that one can check that $lk_\mathbb{Q}(K_1, K_2) \in \mathbb{Q}/\mathbb{Z}$ only depends on the homology classes of $K_1$ and $K_2$.

I hope that now it is clear why there is a bilinear (non-singular) pairing $$lk_\mathbb{Q}: \text{Tor}H_1(Y, \mathbb{Z}) \times \text{Tor} H_1(Y, \mathbb{Z}) \to \mathbb{Q}/ \mathbb{Z}. $$ Clearly there is no meaningful way to extend this to a non-singular pairing on $H_1(Y, \mathbb{Z})$ since $\mathbb{Q}/\mathbb{Z}$ contains only torsion elements. So, in the case when $H_1(Y, \mathbb{Z})=0$ there is no linking pairing. Here I would like to remark that the analogy with four-manifolds is not that illuminating: the intersection form of a smooth four-manifold recognises its topological type while the linking form can't distinguish $S^3$ from the Poincaré homology sphere (just to give a concrete example).

About your second question: yes, you can define the linking form without referring to surfaces. This is a generalisation of the first definition I gave.

$\textbf{Definition 1'.}$ Suppose that $K_1, K_2 \subset Y$ are rationally null-homologous. Since the kernel of the inclusion $i_* : H_1(Y- K_2, \mathbb{Q})\to H_1(Y, \mathbb{Q})$ is generated by a meridian $\mu$ of $K_2$ and $i_*[K_1]=0$ in $H_1(Y, \mathbb{Q})$, there exists $q \in \mathbb{Q}$ such that $[K_1]=q [\mu]$ in $H_1(Y-K_2, \mathbb{Q})$. Define $lk_\mathbb{Q}(K_1, K_2)= q$.

For more about this see exercise 4.5.12 at page. 126 of Gompf-Stipsicz's book (the exercise is solved at the end of the book).