Firstly, no plus construction is needed here. The plus construction is used to kill a perfect subgroup of the fundamental group, but $\pi_1(BU)$ is already trivial.

Next, put $\mathcal{V}=\bigoplus_{n=0}^\infty\mathbb{C}$, and equip this with the standard Hermitian inner product. Let $\mathcal{V}_n$ be the obvious copy of $\mathcal{C}^n$ in $\mathcal{V}$. Let $B$ denote the space of subspaces $\mathcal{W}\leq\mathcal{V}\oplus\mathcal{V}$ such that $\mathcal{W}\cap(\mathcal{V}\oplus 0)$ has finite codimension in $\mathcal{W}$, and also the same finite codimension in $\mathcal{V}\oplus 0$. This is the union of subspaces
$$ B(n) = \{\mathcal{W}\in B : \mathcal{V}_n^\perp\oplus 0 \leq\mathcal{W}\leq\mathcal{V}\oplus\mathcal{V}_n, \dim(\mathcal{W}/(\mathcal{V}_n^\perp\oplus 0))=n\}$$
Now $B(n)$ is homeomorphic to the Grassmannian $G_n(\mathbb{C}^{2n})$, and using this we see that $B$ is a model for $BU$.

Now let $E$ denote the space of inner-product preserving linear maps from $\mathcal{V}^k$ to $\mathcal{V}$. This has an evident action of $\Sigma_k$, which is free because inner-product preserving maps are always injective. It is a standard fact that $E$ is also contractible, so it is a model for $E\Sigma_k$.

Now suppose we have elements $\mathcal{W}_1,\dotsc,\mathcal{W}_k\in B$ and a map $f\in E$. We then have a subspace
$$\bigoplus_i\mathcal{W}_i\leq\bigoplus_{i=1}^k (\mathcal{V}\oplus\mathcal{V})
\simeq \left(\bigoplus_{i=1}^k\mathcal{V}\right)\oplus
\left(\bigoplus_{i=1}^k\mathcal{V}\right).
$$
We can apply $f\oplus f$ to this to get a point $\gamma(f;\mathcal{W}_1,\dotsc,\mathcal{W}_k)\in B$. This construction gives the map $\gamma:E\times_{\Sigma_k}B^k\to B$ that you need.