Goldblatt in "Topoi" defines the relation ≤ on an NNO $N$ as the subobject $\langle \pi_1, \oplus \rangle : N \times N \rightarrowtail N \times N$, where $\oplus : N \times N \rightarrow N$ is the "addition". So far so good — its $\mathbf{Set}$ interpretation is the (sub)set $\{ (m, m + n) \mid m, n \in \mathbb{N} \}$.
Then, he notes that $m < n \iff m + 1 \leq n$, so he defines the relation $<$ as $\leq \circ\ (s \times 1_N)$, where $s$ is the NNO's successor function. This also makes sense.
Or, it made sense until I started doing exercises. In particular, it'd be reasonable to expect, $\mathbf{Set}$-speaking, $0 < 1$. Or, categorically, it means that there exists an arrow $g : 1 \rightarrow N \times N$ such that $\langle 0, 1 \rangle = \text{<} \circ\ g$ (where $0 = O : 1 \rightarrow N$ the "zero" of the NNO, and $1 = s 0$). Clearly $g = \langle a, b \rangle$ for some $a, b$. So we need $$ \langle 0, 1 \rangle =\ < \circ \langle a, b \rangle $$ or (and I'm spelling each step carefully as I might have made a silly mistake somewhere), expanding the definitions, $$ \langle 0, 1 \rangle = \langle \pi_1, \oplus \rangle \circ s \times 1_N \circ \langle a, b \rangle $$ or, collapsing the righmost composition, $$ \langle 0, 1 \rangle = \langle \pi_1, \oplus \rangle \circ \langle s \circ a, b \rangle $$ or, using $\langle f, g \rangle \circ h = \langle f \circ h, g \circ h \rangle$, $$ \langle 0, 1 \rangle = \langle s \circ a, \oplus \circ \langle s \circ a, b \rangle \rangle $$
Now, $\langle f, g \rangle = \langle f', g' \rangle$ implies in particular $f = f'$, so we get $0 = s \circ a$ for some $a$, which clearly cannot happen because Peano axioms and all that stuff.
What's wrong with the above considerations? It feels like I'm missing something terribly trivial, but I can't put my finger on it.
Also it seems like defining ≤ as $\langle \pi_2, \oplus \rangle$ (which is equivalent to $\{ (n, m + n) \mid m, n \in \mathbb{N} \}$ and should perhaps turn out to be an isomorphic subobject in arbitrary topoi as well) sidesteps this problem — the successor gets applied to the first component, but we project out the second so we don't care. This is also concerning as isomorphic subobjects should have the same members after all (at least in $\mathbf{Set}$).
I think it just needs to be $1_N \times s$ instead of $s \times 1_N$, since $m < m + Sn$ for all naturals $n$. Might just be a typo.
Following the same path as you, we want
$$ \langle 0, 1 \rangle = \langle \pi_1, \oplus \rangle \circ 1_N \times s \circ \langle a, b \rangle $$
And then
$$ \langle 0, 1 \rangle = \langle \pi_1, \oplus \rangle \circ \langle a, s \circ b \rangle $$
followed by
$$ \langle 0, 1 \rangle = \langle a, \oplus \circ \langle a, s \circ b \rangle \rangle $$
This is then satisfied with $a = b = 0$.
There's a problem with this line of thinking, which I think is what led to your original confusion. The relation $<$ as defined here should be the subset $\{ (m, n) : m < n\}$ of $N \times N$. By the remark quoted above, this is the subset $\{ (m, n) : Sm \leq n\}$. However, what $\leq \circ\ (s \times id_N)$ defines is not this subset, but rather $\{(Sm, Sm + n) : m, n \in N\}$, which is an entirely different relation. Basically, precomposing with $s$ increments all of the $m$s, not just the one we wanted. As you demonstrated, $0$ can't be related to anything in this relation. Other than that, it's the same as $\leq$.
Let's step back a bit. We have a monomorphism $\leq : N \times N \to N \times N$ and another morphism $s \times id_N : N \times N \to N \times N$. We want to combine these to get a new monomorphism $< : N\times N \to N \times N$ which represents the subset of elements x where $(s \times id_N)(x)$ is in the subset $\leq$. Since all the objects are the same, it's easy to get excited and just compose things, but that won't work out here. Let's generalize so that the objects are all different.
Say we have a subobject $i : A \to Y$ and a morphism $f : X \to Y$. We want to define the subobject roughly equal to $\{x \in X : f(x) \in A\}$, which is the preimage $f^{-1}(A)$. Note that we can't compose these morphisms together, which is a strong hint that the original idea was flawed.
To define the preimage categorically, we use the pullback of $i$ and $f$ (specifically the projection $A \times_Y X \to X$ is the monomorphism we need).
For our situation, this would be the pullback of $\leq$ and $(s \times id_N)$. For sets, this is the set of pairs of pairs $\{((a, b), (m, n)) : (a, a + b) = (S m, n)\}$ and the projection we want is the pair $(m, n)$. Let's simplify and make sure this makes sense.
From the equation (a, a + b) = (S m, n), we see that $a = S m$ and $n = a + b$, so we can simplify $(m, n)$ to $(m, a + b) = (m, S m + b)$.
So the subset of $N \times N$ that we get is $\{(m, S m + b) : m, b \in N\}$, which you can check does contain $(0, 1)$ and all the other pairs you'd expect.
My suggestion at the top of the answer uses the set of pairs $(m, m + S n)$, which amounts to the same thing.