Basically, the norm of the secret key does not depend on $q$, while the norm of the second minimum of the lattice does, so, when you increase $q$, you increase the gap between the first and the second minima of the lattice, which makes the problem easier.
For instance, if the secret key is composed by two degree-$(N-1)$ polynomials $f$ and $g$ with coefficients in $\{-1, 0, 1\}$, then we have $||(f, g)|| < 2N$.
One basis of the lattice $L$ that we use in attack is
$$
\begin{pmatrix}
I_N & H \\
0 & q\cdot I_N
\end{pmatrix}
\in \mathbb{Z}^{2N\times 2N}
$$
So, it is clear that $\det(L) = q^N$ and $\dim(L) = 2N$.
Then, by the Gaussian Heuristic, we expect the short vectors of $L$ to have norm close to
$\sqrt{N}(\det(L))^{1/\dim(L)} = \sqrt{Nq}$.
Because $(f, g) \in L$ and $||(f, g)||$ is smaller than the value predicted by the Gaussian Heuristic, there is a "gap" in $L$, that is, we expect that
$\lambda_1(L) = ||(f, g)||$ and
$\lambda_2(L) \approx \sqrt{Nq}$.
Now, roughly speaking, if the vector that we recovered is shorter than the second shortest vector, then it has to be the first shortest vector.
So, when we run a lattice basis reduction, if we recover a vector smaller than
$\sqrt{Nq}$, it will already be (a short multiple of) $(f, g)$ and we win.