As Max points out in his answer, one is just looking for a complete list of (representatives for) elements of the projective line $\mathbf{P}^1(\mathbf{Z}/N\mathbf{Z})$. Here are some maybe useful observations:

(*) If $\gcd(N,M) = 1$ the natural map $\phi:\mathbf{P}^1(\mathbf{Z}/NM\mathbf{Z}) \to \mathbf{P}^1(\mathbf{Z}/N\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/M\mathbf{Z})$ is bijective.

If one has an element of $([a:b],[c:d]) \in \mathbf{P}^1(\mathbf{Z}/N\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/M\mathbf{Z})$ in order to find $[e:f] = \phi^{-1}(([a:b],[c:d])$ one just needs to solve the congruences $e \equiv a \pmod N$, $e \equiv c \pmod M$, $f \equiv b \pmod N$ and $f \equiv d \pmod M$ for $e,f \in \mathbf{Z}$.

This reduces the problem to the case where $N = p^\ell$ is a power of a prime number $p$. Here one observes:

(**) $|\mathbf{P}^1(\mathbf{Z}/p^\ell\mathbf{Z})| = (p+1)p^{\ell-1}$, and a complete list of representatives is as follows: {$[1:x] \mid x \in p\mathbf{Z}/{p^\ell}\mathbf{Z}$} together with {$[(a + p^\ell\mathbf{Z})+y:1] \mid 0 \le a < p, y \in p\mathbf{Z}/{p^\ell}\mathbf{Z}$}.

From this point of view, one sees the problem(s) with the original list of representatives provided by the OP; e.g., in the list (1,1),(2,1),...,(12,1),(1,2),(3,2),(5,2),(1,3),(2,3),(4,3),(1,4),(3,4),(1,6),(1,12) there are only three pairs (e,f) for which

$$\phi([e:f]) = ([1:1],*) \in \mathbf{P}^1(\mathbf{Z}/4\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/3\mathbf{Z})$$

-- namely (1,1), (5,1), and (9,1) -- and only three pairs (e,f) for which

$$\phi([e:f]) = ([1:0],*) \in \mathbf{P}^1(\mathbf{Z}/4\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/3\mathbf{Z})$$

-- namely (1,4), (3,4) and (1,12) -- and in each case there should be 4 = $|\mathbf{P}^1(\mathbf{Z}/3)|$. Max listed the missing elements, of course.

As Max points out in his answer, one is just looking for a complete list of (representatives for) elements of the projective line $\mathbf{P}^1(\mathbf{Z}/N\mathbf{Z})$. Here are some maybe useful observations:

(*) If $\gcd(N,M) = 1$ the natural map $\phi:\mathbf{P}^1(\mathbf{Z}/NM\mathbf{Z}) \to \mathbf{P}^1(\mathbf{Z}/N\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/M\mathbf{Z})$ is bijective.

If one has an element of $([a:b],[c:d]) \in \mathbf{P}^1(\mathbf{Z}/N\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/M\mathbf{Z})$ in order to find $[e:f] = \phi^{-1}(([a:b],[c:d])$ one just needs to solve the congruences $e \equiv a \pmod N$, $e \equiv c \pmod M$, $f \equiv b \pmod N$ and $f \equiv d \pmod M$ for $e,f \in \mathbf{Z}$.

This reduces the problem to the case where $N = p^\ell$ is a power of a prime number $p$. Here one observes:

(**) $|\mathbf{P}^1(\mathbf{Z}/p^\ell\mathbf{Z})| = (p+1)p^{\ell-1}$, and a complete list of representatives is as follows: {$[1:x] \mid x \in p\mathbf{Z}/{p^\ell}\mathbf{Z}$} together with {$[(a + p^\ell\mathbf{Z})+y:1] \mid 0 \le a < p, y \in p\mathbf{Z}/{p^\ell}\mathbf{Z}$}.

From this point of view, one sees the problem(s) with the original list of representatives provided by the OP; e.g., in the list (1,1),(2,1),...,(12,1),(1,2),(3,2),(5,2),(1,3),(2,3),(4,3),(1,4),(3,4),(1,6),(1,12) there are only three pairs (e,f) for which

$$\phi([e:f]) = ([1:1],?) \in \mathbf{P}^1(\mathbf{Z}/4\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/3\mathbf{Z})$$

-- namely (1,1), (5,1), and (9,1) -- and only three pairs (e,f) for which

$$\phi([e:f]) = ([1:0],?) \in \mathbf{P}^1(\mathbf{Z}/4\mathbf{Z}) \times \mathbf{P}^1(\mathbf{Z}/3\mathbf{Z})$$

-- namely (1,4), (3,4) and (1,12) -- and in each case there should be 4 = $|\mathbf{P}^1(\mathbf{Z}/3\mathbf{Z})|$ such pairs. Max's answer listed the missing pairs, of course.