Why do we need the local triviality condition when working with vector bundles?
From the perspective of $K$-theory, the point of vector bundles is to be able to classify them. That is, given a space $X$, we want a nice classification of all vector bundles (up to isomorphism, or up to stable equivalence) which can both provide a useful invariant of $X$ and give us information about specific naturally occurring vector bundles we might care about.
The local triviality condition makes classifying vector bundles tractible and have deep relationships to other natural questions in topology. Very crucially, the local triviality condition makes it possible to prove that vector bundles are homotopy invariant (at least assuming all our spaces are paracompact): that is, given a vector bundle on $X\times[0,1]$, the two vector bundles on $X$ you get by restricting to $X\times\{0\}$ and $X\times\{1\}$ are isomorphic. In particular, then, a vector bundle on a contractible space is trivial, so if we cover our space by contractible open sets, we can classify vector bundles by thinking about the possible ways to glue together trivial bundles over those open sets via transition functions. This also makes it possible to classify vector bundles in terms of homotopy classes of maps into Grassmannians, which opens up all sorts of connections to homotopy theory such as Bott periodicity.
If you don't require local triviality, then classifying vector bundles is immensely more complicated and intricately related to the exact topology of your space, rather than just its homotopy type. In particular, for instance, if you partition your space $X$ into arbitrary subsets, you could take any vector bundles of the same rank over each of those subsets and just take their disjoint union to get a vector bundle on $X$. There are many more complicated examples. Basically, classifying vector bundles becomes an incredibly complicated pointset topology question and completely useless for any sort of computable invariant.
In my view the condition is necessary in order to verify that we are paramaterizing a family of vector spaces continuously. The situation you describe with one fiber is actually not continuous, and you can check that a nonzero (local) section of the vector bundle would need to "jump" values. This escapes the intuition one want for a family of vector spaces, and in fact makes no use of the topology in the base space, and would work over a set.
The idea of locally trivial verifies that everything varies continuously, and the local trivializations tell us how to glue these locally defined families together.
You may of course consider bundles of a very general type. See for example Chapter 2 of
Husemoller, Dale. Fibre bundles. Vol. 5. New York: McGraw-Hill, 1966.
Husemoller also performs some general contructions with bundles, e.g. products. In that sense you are right.
If you come to vector bundles, you will of course require that all fibres $p^{-1}(x)$ are topological vector spaces. Now consider the projection $p : E = B \times \mathbb R^n \to B$. For each $x \in B$ choose a subspace $V_x \subset \mathbb R^n$. We do not make any assumption concerning the dimension of $V_x$. Then $E' = \bigcup_{x \in B} \{x\} \times V_x$ is a subspace of $E$ and $p$ restricts to $p' : E' \to B$. This would be a vector bundle in your sense. It may look completely erratic. A classification of these bundles up to isomorphism is practically impossible, and there is certainly no connection with the space $B$. In fact, the topology of $B$ does not play any role for the choice function $x \mapsto V_x$.
I would even say that your general bundles are in effect bundles of sets $E, B$. The situation is similar for functions between spaces: You may consider arbitrary functions or continuous functions, but obviously you cannot say much about general functions. Only restriction to smaller classes of functions will produce something interesting. Another example are groups: You cannot say much about general groups, but if you restrict to special classes like finitely generated abelian groups, you get very interesting results.
Thus we should "choose wisely" which type of objects we want to consider. In my opinion general vector bundles would not be a wise choice.
Moreover, many "naturally occurring" vector bundles are locally trivial. Examples are tangent bundles of smooth manifolds.
Two final remarks:
Your example using different orientations does not work. An orientation is an additional structure on a vector space.
"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." Try to define what a continuous parametrization should be. See Andres Mejia's answer.
Edited:
The present question motivated me to think about the meaning of "the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$". I posted a new question Alternative characterization of local triviality of vector bundles?
This contains a suggestion what a continuous parametrization could be. If you accept the corresponding definition, you will see that local triviality is a consequence of being continuously parametrized.