The quotient $\Gamma_0(N) \backslash \mathrm{SL}_2(\mathbf{Z})$ is in bijection with the set of pairs $(c,d)$ where $c$ is a positive divisor of $N$ and $1 \leq d \leq \frac{N}{c}$ satisfies $\operatorname{gcd}(d,c,\frac{N}{c})=1$.

The class of a matrix $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in \mathrm{SL}_2(\mathbf{Z})$ is mapped to $(c,d):=(\operatorname{gcd}(\gamma,N), \delta \mod{\frac{N}{c}})$. It is a good exercice to check that this gives a well-defined and bijective map.

To get an explicit matrix representative associated to $(c,d)$, first find an integer $1 \leq \delta \leq N$ such that $\delta \equiv d \mod{\frac{N}{c}}$ and $\operatorname{gcd}(\delta,c)=1$, then find integers $\alpha,\beta \in \mathbf{Z}$ such that $\alpha \delta- \beta c=1$, and you get your representative $\begin{pmatrix} \alpha & \beta \\ c & \delta \end{pmatrix}$.

(2022.03.27, edited after Tireless and hardworking's remark.)

The quotient $\Gamma_0(N) \backslash \mathrm{SL}_2(\mathbf{Z})$ (right cosets) is in bijection with the set of pairs $(c,d)$ where $c$ is a positive divisor of $N$ and $1 \leq d \leq \frac{N}{c}$ satisfies $\operatorname{gcd}(d,c,\frac{N}{c})=1$.

To get the image of a matrix $M = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in \mathrm{SL}_2(\mathbf{Z})$, first find $u \in (\mathbf{Z}/N\mathbf{Z})^\times$ such that $\gamma u \equiv \operatorname{gcd}(\gamma,N) \bmod{N}$. Then $M$ is mapped to $(c,d):=(\operatorname{gcd}(\gamma,N), \delta u \bmod{\frac{N}{c}})$. It is a good exercice to check that this gives a well-defined and bijective map.

To get an explicit matrix representative associated to $(c,d)$, first find an integer $1 \leq \delta \leq N$ such that $\delta \equiv d \mod{\frac{N}{c}}$ and $\operatorname{gcd}(\delta,c)=1$, then find integers $\alpha,\beta \in \mathbf{Z}$ such that $\alpha \delta- \beta c=1$, and you get your representative $\begin{pmatrix} \alpha & \beta \\ c & \delta \end{pmatrix}$.