Why is topological K-theory equivalent to nonabelian cohomology with respect to the stable unitary group?

Question

I was reading on the $n$Lab page for topological K-theory that taking cohomology of a smooth space with respect to the smooth $\infty$-stack $\mathbf{Vect}$ is equivalent to taking its cohomology with respect to the stable unitary group, and it is not obvious to me as to why this is. Is it simply because any map into $\mathbf{B}U$ has to come from a map into $\mathbf{B}U(n)$ for some $n$ since it is a homotopy colimit over these classifying spaces?