Fix $ B = (-1,1) $ to be the base space, and to each point $ b $ of $ B $, attach the vector-space fiber $ \mathcal{F}_{b} \stackrel{\text{def}}{=} \{ b \} \times \mathbb{R} $. We thus obtain a trivial $ 1 $-dimensional vector bundle over $ B $, namely $ B \times \mathbb{R} $. Next, define a fiber-preserving vector-bundle map $ \phi: B \times \mathbb{R} \rightarrow B \times \mathbb{R} $ as follows: $ \forall (b,r) \in B \times \mathbb{R}: \quad \phi(b,r) \stackrel{\text{def}}{=} (b,br). $ We now consider the kernel $ \ker(\phi) $ of $ \phi $. For each $ b \in B $, let $ \phi_{b}: \mathcal{F}_{b} \rightarrow \mathcal{F}_{b} $ denote the restriction of $ \phi $ to the fiber $ \mathcal{F}_{b} $. Then $ \ker(\phi_{b}) $ is $ 0 $-dimensional for all $ b \in (-1,1) \setminus \{ 0 \} $ but is $ 1 $-dimensional for $ b = 0 $. Hence, $ \ker(\phi) $ does not have a local trivialization at $ b = 0 $, which means that it is not a vector bundle.
In general, if $ f: \xi \rightarrow \eta $ is a map between vector bundles $ \xi $ and $ \eta $, then $ \ker(f) $ is a sub-bundle of $ \xi $ if and only if the dimensions of the fibers of $ \ker(f) $ are locally constant. It is also true that $ \text{im}(f) $ is a sub-bundle of $ \eta $ if and only if the dimensions of the fibers of $ \text{im}(f) $ are locally constant.
The moral of the story is that although something may look like a vector bundle by virtue of having a vector space attached to each point of the base space, it may fail to be a vector bundle in the end because the local trivialization property is not satisfied at some point. You want the dimensions of the fibers to stay locally constant; you do not want them to jump.
Richard G. Swan has a beautiful paper entitled Vector Bundles and Projective Modules (Transactions of the A.M.S., Vol. 105, No. 2, Nov. 1962) that contains results that might be of interest to you.