I was trying to figure out the homology of the knot complement when $K$ is not a rational null-homologous knot ($[K]\neq 0\in H_1(X_K,\mathbb Q)$).
We then know by half-lives half dies that the meridian class has to be a torsion element or zero in the complement. By using the Mayer-Vietoris sequence with the knot neighbourhood and the complement I get:
$$0\rightarrow \mathbb Z/k\mathbb Z[u]\oplus\mathbb Z[l]\rightarrow H_1(X_K)\oplus\mathbb Z[l]\rightarrow H_1(Y)\rightarrow 0$$
then I do not know how to keep going.
Obviously $[K]\neq 0$ in the complement. Hence the image in Mayer Vietoris identifies this element $[K]$ with $l$ to obtain $H_1Y$ (to compute the latter is our goal for now).
As you mentioned correctly, the meridian has to be (possibly trivial) torsion in $H_1X_K$. Anyway we obtain $H_1Y$ as the quotient $H_1X_K \oplus \mathbb Z / (([K],1),([M],0)) \cong H_1X_K$, where the last isomorphism holds iff $[M]=0$.
To show that this is in fact always true (at least in the orientable case), and hence our result is $H_1X_K\cong H_1Y$, note that by Poincaré duality we can find an orientable surface $S$ in $Y$ intersecting transversely $K$ precisely once. Now $X_K\cap S$ gives us a nullbordism.
(or in fancier words: in the exact Mayer Vietoris sequence above the meridian is in the image of $H_2Y$ and hence maps to zero by exactness)