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.
First of all, the two vector bundles could have different rank (in which case they could not possibly be isomorphic, but there could still be non-trivial maps between them), i.e., the $n$ of $p^{-1}(b) \cong \{b\} \times \mathbb{R}^n$ does not have to be the same as the $n'$ of $p'^{-1}(b) \cong \{b\} \times \mathbb{R}^{n'}$. And secondly, even if $n=n'$, not every map from $\mathbb{R}^n$ to $\mathbb{R}^n$ is an isomorphism. You are not just requiring that the fibers are isomorphic in some abstract way, but that the unique map induced by the morphism of vector bundles is an isomorphism. And this is far from being automatic. To make it more explicit, you have the following commutative diagram:
$$\require{AMScd}
\begin{CD}
p^{-1}(\{b\}) @>{F}>> p'^{-1}(\{b\})\\
@V{\varphi}VV @V{\varphi'}VV \\
\{b\} \times \mathbb{R}^n @>{\varphi' \circ F \circ \varphi^{-1}}>> \{b\} \times \mathbb{R}^{n'}
\end{CD}$$
and $F$ is an isomorphism if and only if the bottom map is an isomorphism, but this is not automatic from the fact that $\varphi$ and $\varphi'$ are isomorphisms (where $\varphi:= \varphi_\alpha\vert_{p^{-1}(\{b\})}$ for some chart $U_\alpha$ containing $p^{-1}(\{b\})$, and similarly for $\varphi'$).
Best Answer
first of all if you have maps $f:X\to S$,$g:Y\to S$,by a $S$ morphism between $h:X\to Y$ we mean that $f=h\circ g$(so the two way you can go from $X$ to $S$ should be the Same). In your example you have maps $E_{|U}\to U$,$U\times V\to U$ so you can talk about a $U$ morphism between $E_{|U}$ and $U\times V$(the point of a $S$ morphism $h$ is basically that you want $h$ to sends fiber over $s$ in $X$ to the fiber over $s$ in $Y$ ).
When you define a vector space, you don't mention continuity because you don't even have a topology, but it is an easy exercise that if you have a topological field(like $\mathbb{R},\mathbb{C}$) if you put the product topology on $V$, the sum and scalar product become continuous.
about the relationship between definition and proposition: when you have $p: E\to B$ by definition you have a sum and scalar product on each fiber so you get maps $E\times E\to E$, and $F\times E\to E$ and these are continuous because $E$ locally looks like $U\times V$ and for $U\times V$ you know that these maps are continuous(by previous paragraph).
the idea of the bundle is that it is a space $E$ over $B$(meaning that there is a map $E\to B$) that locally looks like the product of $B$ with a fixed space. if this fixed space is a vector space, we call $E$ a vector bundle. but you can also talk about a circle-bundle or a $F$-bundle for any fixed space you pick.