This question is motivated by Why do we need the local triviality condition when working with vector bundles?
A general bundle is a triple $(E,p,B)$, where $E,B$ are topological spaces and $p : E \to B$ is a map. This concept can be found for example in [1]. A $k$-dimensional pre-vector bundle is defined as a bundle such that each fiber $E_b = p^{-1}(b)$, $b \in B$, has the structure of a $k$-dimensional topological vector space over $\mathbb K$. Atiyah [2] denotes this as a family of vector spaces. A vector bundle is then a locally trivial pre-vector bundle. Most standard constructions for vector bundles also work for pre-vector bundles (direct sum, …).
In the question quoted above the OP states "If the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$, then the local triviality condition shouldn't be necessary." But what should a continuous parameterization be?
In the general case I do not see an obvious definition. So let us look at the following special case. Let $\pi : B \times \mathbb K^n \to B$ be the projection. To each $b \in B$ assign a $k$-dimensional subspace $f(b) \subset \mathbb K^n$. Then $E(f) = \bigcup_{b \in B} \{b\} \times f(b)$ is a subspace of $E$ and $\pi$ restricts to $p : E(f) \to B$. This gives us a $k$-dimensional pre-vector bundle $\xi(f)$ over $B$ which may look completely erratic. A classification of these bundles up to isomorphism is practically impossible, and there is no manifest connection with the topology on $B$.
However, we can regard the above assigment as a function $f : B \to G_k(\mathbb K^n)$ into the Grassmann variety which has a nice metric. It makes sense to regard the function $f$ as the parameterization of $\xi(f)$. Then the parameterization can be defined to be continuous if $f$ is continuous.
Question: What is the relation between "continuously parameterized" and "locally trivial"? Do these concepts agree?
[1] Husemoller, Dale. Fibre bundles. Vol. 5. New York: McGraw-Hill, 1966. See Chapter 2.
[2] Atiyah, Michael. K-theory. CRC Press, 2018.
Best Answer
Moishe Kohan's comment contains the answer of the question. This community wiki elaborates it a little.
Let us first observe that if $\xi = (E,p,B)$ is a pre-vector bundle and $f : X \to B$ is a (not necessarily continuous) function on a space $X$, we get a pullback pre-vector bundle $$f^*(\xi) = (f^*(E), p^*,X)$$ where $$f^*(E) = \bigcup_{x \in X} \{x \} \times p^{-1}(f(x)) \subset X \times E$$ and $p^*$ is the restriction of the projection $X \times E \to X$.
Now we generalize the construction of the question (see [1]). Define $$\mathbb K^\infty = \{(x_i)_{i \in \mathbb N} \mid x_i \in \mathbb K, x_i = 0 \text{ for almost all } i \} .$$ This is a vector space with an obvious inner product and we may regard each $\mathbb K^n$ as a genuine subspace of $\mathbb K^\infty$. Doing so, we have $\mathbb K^\infty = \bigcup_{n \in \mathbb N} \mathbb K^n$.
For $0 \le m \le \infty$ and $k \in \mathbb N$ let $G_k(\mathbb K^m)$ denote the set of all $k$-dimensional linear subspaces of $\mathbb K^m$. For $m < \infty$ these are the well-known Grassmann varieties. Each $G_k(\mathbb K^n)$ is a genuine subspace of $G_k(\mathbb K^{n+1})$, and we define $G_k(\mathbb K^\infty) = \bigcup_{n \ge k} G_k(\mathbb K^n)$ as a set, and $U \subset G_k(\mathbb K^\infty)$ to be open iff $U \cap G_k(\mathbb K^n)$ is open in $G_k(\mathbb K^n)$ for all $n$. Thus $G_k(\mathbb K^\infty)$ is the direct limit of the sequence of spaces $G_k(\mathbb K^n)$ bonded by inclusions.
The tautological (or canonical bundle) $\gamma^m_k$ over $G_k(\mathbb K^m)$ has as total space $$E^m_k = \bigcup_{V \in G_k(\mathbb K^m)} \{V\} \times V \subset G_k(\mathbb K^m) \times \mathbb K^m$$ with obvious projection $\pi$ onto the base. The fiber over $V \in G_k(\mathbb K^m)$ is nothing else than $\{V\} \times V$, i.e. a copy of $V \subset \mathbb K^m$. Again we have $E_k^\infty = \bigcup_{n \ge k} E_k^n$. It is well-known that $\gamma^m_k$ is locally trivial.
Now let $f : B \to G_k(\mathbb K^m)$ be any (not necessarily continuous) function. The pullback pre-vector bundle $f^*(\gamma^m_k)$ over $B$ has as total space $$f^*(E_k^m) = \bigcup_{b \in B} \{b \} \times \pi^{-1}(f(b)) = \bigcup_{b \in B} \{b \} \times \{f(b)\} \times f(b) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m .$$ In general this is completely erratic. Let us say that $f^*(\gamma^m_k)$ is continuously parameterized if $f$ is continuous.
Note, however, that $f^*(\gamma^m_k)$ is not the same pre-vector bundle as $\xi(f)$ which was defined in the question. But $f^*(\gamma^m_k)$ is continuously parameterized if and and only $\xi(f)$ is. It seems that we now habe an adequate interpretation of "the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$", at least for pre-vector bundles of the form $f^*(\gamma^m_k)$ and $\xi(f)$.
Fact 1. If $f^*(\gamma^m_k)$ is continuously parameterized, then it is locally trivial.
This is well-known. Pullbacks of vector bundles along continuous maps are always vector bundles. This shows that being continuously parameterized is even stronger than locally trivial.
Fact 2. If $f^*(\gamma^m_k)$ is locally trivial, then it is continuously parameterized.
Let $s_0 : B \to f^*(E_k^m) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m$ be the zero-section which is given $s_0(b) = (b,f(b),0)$. Each $b \in B$ has an open neighborhood $U \subset B$ such that the restriction of $f^*(\gamma^m_k)$ to $U$ is trivial. This implies that $s_0 \mid_U$ is continuous. Since the projection $p_2 :B \times G_k(\mathbb K^m) \times \mathbb K^m \to G_k(\mathbb K^m)$ is continuous, we see that $f \mid_U = p_2 \circ s_0 \mid_U$ is continuous. Thus $f$ is continuous.
Fact 3. If $f$ is continuous, then $f^*(\gamma^m_k)$ and $\xi(f)$ are isomorphic. In particular, $\xi(f)$ is locally trivial.
To see this, define $\phi_f : B \times \mathbb K^m \to B \times G_k(\mathbb K^m) \times \mathbb K^m, \phi(b,v) = (b,f(b),v)$. We have $\phi_f(E(f)) = f^*(E_k^m)$ so that we get an induced $\phi'_f : E(f) \to f^*(E_k^m)$ which is bijection which maps the fiber over $b$ in $E(f)$ by a linear isomorphism to the fiber over $b$ in $f^*(E_k^m)$. It is continuous iff $f$ is continuous. Next define $\psi : B \times G_k(\mathbb K^m) \times \mathbb K^m \to B \times \mathbb K^m$ as the projection. Clearly $\psi$ is continuous and $\psi(f^*(E_k^m)) = E(f)$. Hence the restriction $\psi_f : f^*(E_k^m) \to E(f)$ is a morphism of pre-vector bundles which is continuous and fiberwise linearly isomorphic. It is an isomorphism of pre-vector bundles iff $f$ is continuous.