Is this a bundle? In Frederic Schuller's lecture series Lectures on The Geometrical Anatomy of Theoretical Physics, he gives an example of a bundle $E\overset{\pi}{\rightarrow}M$ where different points of the base manifold have different fibres:

Or, as spelled out in Simon Rea's transcription of the lectures, found here (p. 39):
$$F_p:=\mathrm{preim}_\pi(\{p\}) \cong_\mathrm{top}
\begin{cases} 
 S^{1} & p < 0\\
 \{ p \} & p = 0\\
 [0,1] & p > 0\\
\end{cases}$$
The example is great for intuition. But I don't see how E can be a manifold with such an odd structure (as a reminder, the definition of $E\overset{\pi}{\rightarrow}M$ requires E to be a topological manifold).
 A: Indeed, the total space here fails to be a manifold -- more precisely, it's impossible to have any continuous surjection $E \to \mathbb{R}$ from a topological manifold $E$ with the prescribed preimages. To see this, note that removing the unique point in the preimage of $0$ disconnects $E$, so if $E$ were a manifold, it would have to be $1$-dimensional. However, $E$ cannot be a $1$-dimensional manifold, as follows.
It is a standard result that every non-compact, connected topological $1$-manifold is homeomorphic to $[0,\infty)$ or $\mathbb{R}$. There cannot be a continuous surjection from one of these spaces to $\mathbb{R}$ with the prescribed fibers, since $S^1$ is not homeomorphic to a subspace of $\mathbb{R}$ (every connected compact subspace of $\mathbb{R}$ is of the form $[a,b]$ with $a \leq b \in \mathbb{R}$, and these subspaces are simply connected while $S^1$ is not).

From a brief watch of a few minutes, it seems like these lectures have a few other problems with mathematical precision.
For example, the definition of "fiber bundle" given is incorrect, or at the very least extremely nonstandard. Please compare to the definition on Wikipedia, and see this post which may be referencing these same lectures.
Also, the lecturer claims that the Möbius strip is homeomorphic to the cylinder, which is simply not true (the Möbius strip is not homeomorphic to a subspace of $\mathbb{R}^2$, while the cylinder is).
So, I would take the mathematical details in these videos with a grain of salt (perhaps consider studying alongside a reputable math textbook on topological manifolds).
A: It is not a bundle of topological manifolds, but it is a bundle of topological spaces.
If you want, you can remove the problematic point both in $E$ and in $M$, giving you something which is a (surjective) bundle of topological manifolds.
I say surjective because there are people for whom "bundles" need not be surjective. With this broader definition, it suffices to remove the point in $E$, and in that way $E$ is a topological manifold, the only thing is that the fibre over $q$ is now empty.
If you want a simple example of a surjective bundle over $\mathbb R$ which is not a fibre bundle, you can consider the map $\mathbb R\to\mathbb R$ given like this:
