Let $E_1, E_0$ be vector bundles over a manifold $X$. Let us suppose that $$E_1 - E_0 =0 \in K(X)$$ (I believe we also suppose $X$ to be compact so $K$-theory makes sense here.)
Proposition: If $\dim E_i > \dim X$, then there is an isomorphism, $$ E_1 \rightarrow E_0$$ over $X$.
I am greatful for references or proof. I have not found any material regarding bundles over manifolds.
Source: This is claimed in page 20 line 12 of this paper by Atiyah.
This follows from the existence of classifying spaces. First let me work with reduced K-theory, as if $E_1 = E_2$ in $K$-theory itself, then the rank of $E_1$ and $E_2$ are the same, and conversely if $E_1$ and $E_2$ have the same rank and they are equal in reduced $K$-theory, then they are equal in $K$-theory.
We have $$\tilde K^0(X) = [X, BO],$$ while $$\text{Vect}_n(X) = [X, BO(n)].$$
The map $\text{Vect}_n(X) \to \tilde K^0(X)$ is given by sending $BO(n) \hookrightarrow BO$.
Proof: there is a fibration $S^n \to BO(n) \to BO(n+1)$, which induces an exact sequence on mapping sets $[X, S^n] \to [X, BO(n)] \to [X, BO(n+1)]$; because $[X, S^n] = 0$ as $n > \dim X$, the map $\text{Vect}_n(X) \to \text{Vect}_{n+1}(X)$ is injective. For surjectivity (which applies one dimension lower), we need to show that any bundle of rank $n > \dim X$ has a trivial line bundle as summand. Equivalently, we need to show that every sphere bundle $S^{n-1} \hookrightarrow E \to X$ has a section as soon as $n > \dim X$. This follows from obstruction theory, which identifies the obstruction to the existence of a section as an element of the set $H^n(X;\pi_{n-1} S^{n-1}) = 0$, by the assumption that $n > \dim X$.
Because $BO = \text{colim }BO(n)$, you have (so long as your spaces aren't silly) an equality $$\tilde K^0(X) = [X, BO] = \text{colim } [X, BO(n)] = \text{colim Vect}_n(X).$$
By the lemma, so long as $n > \dim X$, the map $\text{Vect}_n(X) \to \tilde K^0(X)$ is a bijection. This is what you wanted.