Let $\Omega = (\mathbb{N}_{\infty} \xrightarrow{p} \mathbb{N}_{\infty} \xrightarrow{p} \dotsc)$ be as described.
Let $S \subseteq X$ be a subobject, thus we have a bunch of compatible injections $S_i \to X_i$. Compatibility means that the diagrams
$$\begin{array}{c} X_i & \rightarrow & X_{i+1} \\ \downarrow && \downarrow \\ S_i & \rightarrow & S_{i+1} \end{array}$$
commute.
Define $\phi : X \to \Omega$ as follows: If $i \in \mathbb{N}$, we want to define $\phi_i : X_i \to \Omega_i = \mathbb{N}_{\infty}$. Well, if $x \in X_i$, then there are three cases:
$x \in S_i$ (by which I mean that $x$ lies in the image of $S_i \to X_i$). Then $\phi_i(x):=0$.
More generally, assume that the image of $x$ in $X_{i+n}$ lies in $S_{i+n}$ for some $n \geq 0$. Choose $n$ minimal. Then $\phi_i(x) := n$.
Otherwise, we define $\phi_i(x) := \infty$.
By the very construction, the diagram
$$\begin{array}{c} X_i & \rightarrow & X_{i+1} \\ \phi_i \downarrow ~~~~ && ~~~~ \downarrow \phi_{i+1} \\ \mathbb{N}_\infty & \xrightarrow{p} & \mathbb{N}_\infty \end{array}$$
commutes, i.e. $\phi : X \to \Omega$ is a morphism. One can also check that we have a pullback diagram, as desired.
Rather than requiring $\mathbb{C}$ to be a locally small category with pullbacks, the correct requirement here is actually that $\mathbb{C}$ be small. If $\mathbb{C}$ is not small, then we should not expect $\Omega(A)$ to form a set as defined.
For instance, let $\mathbb{C}$ be the ordinals, considered as a category under the reverse order. Then $\mathbb{C}$ has pullbacks and is locally small. But each ordinal $\beta \geq \alpha$ gives rise to a sieve $S_\beta = \{\gamma \mid \gamma \geq \beta\}$ on $\alpha$, and $S$ is injective, so the class of sieves on $\alpha$ is a proper class and not a set. So $\Omega$ is not well-defined in this instance.
Let's move on to the substance of the question. We start out by supposing that $\sigma$ is the pullback of $\top$ along some morphism $\chi^\sigma$.
Recall that a diagram is a pullback square in $\hat{\mathbb{C}}$ if and only if the diagram, evaluated at each $A \in \mathbb{C}$, is a pullback square in $\mathbb{C}$. So we have, for each $A$, a pullback diagram as pictured in the question.
Let's begin by noting that
Lemma: For all $A \in \mathbb{C}$, for all $x \in X(a)$, $\downarrow A = \chi^\sigma_A(x)$ if and only if $x \in F(A)$.
Proof: this follows immediately from the pullback diagram evaluated at $A$. $\square$
Corollary: For all $A \in \mathbb{C}$, for all $x \in X(a)$, $1_A \in \chi^\sigma_A(x)$ if and only if $x \in F(A)$.
Proof: $1_A \in \chi^\sigma_A(x)$ if and only if $\chi^\sigma_A(x) = \downarrow A$ if and only if $x \in F(A)$. $\square$
Let's introduce some notation.
Notation. For some presheave $P$, some objects $A, B \in \mathbb{C}$, some morphism $f : B \to A$, and some $x \in P(A)$, write $x \cdot f = P(f)(x)$. Note that $x \cdot 1_A = x$ and that $x \cdot (f \circ g) = (x \cdot f) \cdot g$. $\square$
Now, we're ready to move on to the major theorem.
Thm. Consider objects $A, B \in \mathbb{C}$, arrow $f : B \to A$, and $x \in X(A)$. Then $f \in \chi^\sigma_A(x)$ if and only if $x \cdot f \in F(A)$.
Proof: note that $\chi^\sigma_B(x \cdot f) = \chi^\sigma_A(x) \cdot f$ by naturality. Now using the definition of $\cdot f$ on $\Omega$, we see that $f \in \chi^\sigma_A(x)$ if and only if $1_B \in \chi^\sigma_A(x) \cdot f$. And of course, $1_B \in \chi^\sigma_A(x) \cdot f = \chi^\sigma_B(x \cdot f)$ if and only if $x \cdot f \in F(B)$ by the corollary. $\square$
Thus, we can write $\chi^\sigma_A(x) = \{f : B \to A \mid x \cdot f \in F(B)\}$, which is exactly what we desired to prove.
Note that what we have shown so far is
If $\sigma$ is the pullback of $\top$ along $\chi^\sigma$, then for all $A$, for all $x \in X(A)$, $\chi^\sigma_A(x) = \{f : B \to A \mid x \cdot f \in F(B)\}$.
What we then need to show is the flip side:
Consider the family of functions defined by $\chi^\sigma_A(x) := \{f
: B \to A \mid x \cdot f \in F(B)\}$. This family is a natural transformation $X \to \Omega$, and the pullback of $\top$ along $\chi^\sigma$ is $\sigma$.
This gives us both existence and uniqueness.
Best Answer
One way to think of a topos is as some kind of fancier category of sets.
One way to make sets fancier is to consider sheaves of sets on some topological space; this is mentioned in Zhen Lin's answer. One can think of a sheaf of sets as a set which varies and twists over the topological space, but it seems that this could be a bit painful for you to think about with the background you're coming from.
Another way to makes sets fancier is to put actions on them. So:
Let $G$ be a finite group, and consider the category of all finite $G$-sets, i.e. all finite sets equipped with an action of the group $G$. (Morphisms are maps between sets that are compatible with the $G$-action on source and target.) This is an example of a topos, which is pretty small in your non-technical sense.
The subobject classifier is the two element set $\Omega$, with trivial $G$-action, and with one of the two points distinguished. If $X$ is a $G$-set, and $Y$ a $G$-invariant subset, then we have the morphism $\chi_Y: X \to \Omega$ which maps all of $Y$ to the distinguished point in $\Omega$, and all of $X\setminus Y$ to the other point. This morphism "classifies" the subset $Y$. (More precisely, $Y$ is the preimage of the distinguished point of $\Omega$ under $\chi_Y$.)
If we take $G$ to be the trivial group, then we just recover the topos of finite sets.
Let's instead take $G$ to be the cyclic group of order two, say $G = \langle 1,\tau\rangle,$ with $\tau$ of order two. Then to give a finite $G$-set is just to give a finite set $X$ equipped with an involution (i.e. permutation of order two) $\tau$.
Now in addition to the two-element set $\Omega$ with trivial $G$-action, which (once you designate one of its points as being the distinguished one) is the subobject classifier, you could think about the two-element set $\Omega'$ equipped with the non-trivial involution, which switches the two points.
We have already seen that $Hom_G(X,\Omega)$ (I write $Hom_G$ for maps preserving the $G$-action) is equal to the collection of $G$-invariant subsets of $X$.
What about $Hom_G(X,\Omega')$? You could try to compute this (of course it will depend on the particular $G$-set $X$). It's not particularly exciting, but it might help you get a feeling for the difference between the subobject classifier and some other objects, such as $\Omega'$.