Fiber products of manifolds Let $\mathsf{Man}$ be the category of smooth manifolds. Denote by $|~|$ the forgetful functor to $\mathsf{Top}$. If $X \to S$ and $Y \to S$ are morphisms in $\mathsf{Man}$, then $X \times_S Y$ exists and $|X \times_S Y| = |X| \times_{|S|} |Y|$ under suitable assumptions, for example when $X \to S$ and $Y \to S$ are submersions, right? But my actual question is:
Question. What is an explicit example for morphisms $X \to S$ and $Y \to S$ such that $X \times_S Y$ does not exist in $\mathsf{Man}$?
Of course it is not enough (a priori) to prove that $|X| \times_{|S|} |Y|$ is not a manifold. On the other hand, it is easy to see, using morphisms from the point, that the underlying set of $X \times_S Y$ is the fiber product of the underlying sets.
I am also interested in related categories, for example topological manifolds, Banach or Frechet manifolds. Also I would like to add a soft question: The theory of manifolds doesn't really suffer from the failure of the existence of fiber products, right? On the other hand fiber products play an essential role in the theory schemes, which can be seen as "algebraic manifolds". What is a possible reason or explanation for this asymmetry?

Here's an argument for the first question. Take $X=Y=S=\mathbb R$ and take $x\mapsto x^2$ for the maps $X,Y\to S$. As you note, the underlying set of $X\times_S Y$ is what you'd expect, $\{(x,y)\mid x=\pm y\}$ (at least up to bijection, and so without loss of generality).
Lemma: Let $Z$ be a subset of $\{(x,y)\mid x=\pm y\}$ such that $Z$ defines a smooth manifold $M$ in the usual topology coming from $X\times Y$. Then the subspace topology induced on $Z$ from $X\times_S Y$ is the usual topology.
Proof: By the universal property of $X\times_S Y$ (and using morphisms from the point), the inclusion $M\to X\times Y$ factorises in $\mathsf{Man}$ into two inclusions:
$$M\to X\times_S Y\to X\times Y$$
Restricting to $Z$ we get continuous inclusions which we can denote $M\to M'\to M$. But this implies $M=M'$.
$\square$
In particular, applying the lemma three times, with $Z$ equal to $\{(x,y)\mid x=\pm y\}\setminus\{(0,0)\}$ and $\{(x,x)\mid x\in\mathbb R\}$ and $\{(x,-x)\mid x\in\mathbb R\}$, we find that $X\times_S Y$ induces the usual topology on these sets. So $X\times_S Y$ is connected but $X\times_S Y\setminus\{(0,0)\}$ has four connected components. This cannot happen with smooth manifolds: removing a point from a manifold increases the number of connected components by at most $1$.
