Showing a subgroup of $\mathbb{Z}\times\mathbb{Z}$ is cyclic.

As you seem to have perceived, this is as much a linear algebra question as a group theory question, although you have to be careful and do your linear algebra over $\mathbb Z$ instead of the usual $\mathbb R$. That means you can only use integers where you are used to using arbitrary real numbers.

The group $H$ is generated by two integer vectors $\vec v = (-5,1)$ and $\vec w = (1,-5)$. Since this is an abelian group, then yes, you can say that $H$ is the group of all integer linear combinations of $\vec v$ and $\vec w$.

Now let's put those two vectors into the rows of a matrix: $$M = \begin{pmatrix} -5 & 1 \\ 1 & -5 \end{pmatrix} $$ It follows that the row space of $M$ over $\mathbb Z$ is $H$, i.e. the set of all integer linear combinations of the rows of $M$ is $H$.

Now use your linear algebra skills to simplify the matrix $M$ by doing row operations which do not affect the row space over $\mathbb Z$. For example, add $5$ times row 2 to row 1 to get $$\begin{pmatrix} 0 & -24 \\ 1 & -5 \end{pmatrix} $$ then switch rows 1 and 2 to get $$\begin{pmatrix} 1 & -5 \\ 0 & -24 \end{pmatrix} $$ and then multiply row $2$ by $-1$ to get $$\begin{pmatrix} 1 & -5 \\ 0 & 24 \end{pmatrix} $$ You can also do column operations over $\mathbb Z$, which have the effect of changing the given basis for $G$, but of course that does not affect the isomorphism type of the quotient group $G/H$. So, adding $5$ times column $1$ to column $2$ you get $$\begin{pmatrix} 1 & 0 \\ 0 & -24 \end{pmatrix} $$ So now we know that $$G / H \approx (\mathbb Z \oplus \mathbb Z) / (\mathbb Z \oplus 24\mathbb Z) \approx (\mathbb Z / 1 \mathbb Z) \oplus (\mathbb Z / 24\mathbb Z) \approx \mathbb Z / 24\mathbb Z $$ so the quotient is isomorphic to the cyclic group of order $24$.