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.

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, note that the space $\mathbb{C}^\infty=\bigoplus_{n=0}^\infty\mathbb{C}$ has a natural Hermitian inner product (with respect to which it is not complete). For any Hermitian space $\mathcal{V}$ that is isomorphic to $\mathbb{C}^\infty$, we consider the space $\mathcal{V}\oplus\mathcal{V}$ and its subspaces $\mathcal{V}_L=\mathcal{V}\oplus 0$ and $\mathcal{V}_R=0\oplus\mathcal{V}$. Let $B(\mathcal{V})$ denote the space of subspaces $\mathcal{A}\leq\mathcal{V}\oplus\mathcal{V}$ such that $\mathcal{A}\cap\mathcal{V}_L$ has finite codimension in $\mathcal{A}$, and also the same finite codimension in $\mathcal{V}_L$. To understand this in more detail, suppose we have a subspace $V\leq\mathcal{V}$ with $\dim(V)=n<\infty$, giving a decomposition $$ \mathcal{V}\oplus\mathcal{V} = V_L\oplus (V^\perp)_L \oplus V_R \oplus (V^\perp)_R. $$ We put $$ B(\mathcal{V};V) = \{A\oplus (V^\perp)_L : A\leq V_L\oplus V_R,\; \dim(W) = n\}. $$ We find that $B(\mathcal{V};V)$ is naturally identified with a finite-dimensional Grassmann manifold, so it has a natural compact Hausdorff topology. Moreover, the set $B(\mathcal{V})$ is the colimit of the sets $B(\mathcal{V};V)$, so we give it the colimit topology. One can check that $B(\mathcal{V})$ is then a model for the homotopy type $BU$.

Now suppose we have two Hermitian spaces $\mathcal{V}$ and $\mathcal{W}$ as above, and a linear map $\alpha\colon\mathcal{V}\to\mathcal{W}$ that preserves inner products. (This implies that $\alpha$ is injective, but it need not be surjective.) Given a point $\mathcal{A}=A\oplus(V^\perp)_L\in B(\mathcal{V};V)$, we have a point $$ \mathcal{B} = (\alpha\oplus\alpha)(A) \oplus(\alpha(V)^\perp)_R \in B(\mathcal{W};\alpha(V)). $$ One can check that this does not really depend on the choice of $V$, so we have a well-defined map $\alpha_*\colon B(\mathcal{V})\to B(\mathcal{W})$. One can also check that this is functorial. There are also evident maps $$ B(\mathcal{V})\times B(\mathcal{W}) \to B(\mathcal{V}\oplus\mathcal{W}), $$ making $B$ into a lax monoidal functor.

Now let $E(k)$ 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 $E(k)$ is also contractible, so it is a model for $E\Sigma_k$.

Now suppose we have elements $\mathcal{A}_1,\dotsc,\mathcal{A}_k\in B(\mathcal{V})$ and a map $\alpha\in E$. We then apply $\alpha_*$ to $\bigoplus_i\mathcal{A}_i$ to get a point $\gamma(\alpha;\mathcal{A}_1,\dotsc,\mathcal{A}_k)\in B(\mathcal{V})$. This construction gives the map $$ \gamma:E(k)\times_{\Sigma_k}B(\mathcal{V})^k\to B(\mathcal{V}) $$ that you need.

Note: An earlier version of this answer said that $\alpha_*(\mathcal{A})$ should just be $(\alpha\oplus\alpha)(\mathcal{A})$, but that is not correct, and in fact $(\alpha\oplus\alpha)(\mathcal{A})$ need not lie in $B(\mathcal{W})$. I thank Jack Smith for pointing out this error.

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 finite codimension in $\mathcal{V}\oplus 0$. This is the union of subspaces $$ B(n) = \{\mathcal{W}\in B : \mathcal{V}_n^\perp\oplus 0 \leq\mathcal{W}\leq\mathcal{V}\oplus\mathcal{V}_n\}$$ Now $B(n)$ is homeomorphic to the Grassmannian $G_n(\mathcal{C}^{2n})$, 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.

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.