Some background before the math:
I am reading a paper that, in its appendix, looks at chemical reactions of the form $$X_i+Y_i\overset{k_i^+}{\underset{k_i^-}{\rightleftharpoons}} Z_i$$ where species $X_i$ and $Y_i$ reversibly combine to make species $Z_i$ with forward rate constant $k_i^+$ and reverse rate constant $k_i^-$. The paper analyzes many coupled reactions of this form, where the $Z_i$ of reaction $i$ could be $X_j$ or even $Z_j$ for some other reaction $j$.
The paper uses the law of mass action to obtain differential equations for the concentrations of the various species: \begin{align} \frac{\text dX_i}{\text dt}=k_i^-Z_i-k_i^+X_iY_i \\\\ \frac{\text dY_i}{\text dt}=k_i^-Z_i-k_i^+X_iY_i \\\\ \frac{\text dZ_i}{\text dt}=k_i^+X_iY_i-k_i^-Z_i \end{align} where coupled reactions are combined into a single differential equation. For example, if we have two reactions $$A_i+B_i\overset{k_i^+}{\underset{k_i^-}{\rightleftharpoons}} C_i$$ $$A_j+B_j\overset{k_j^+}{\underset{k_j^-}{\rightleftharpoons}} C_j$$ and $C_i=C_j$, then we have $$\frac{\text dC_i}{\text dt}=k_i^+A_iB_i+k_j^+A_jB_j-(k_i^-+k_j^-)C_i$$
The paper states at the beginning of section A4:
Using the law of mass action, the kinetic schemes presented above can be formulated in terms of a set of coupled ordinary differential equations. It suffices to give these equations in the shorthand form
$$\frac{\text d\mathbf u}{\text dt}=\mathbf A\mathbf u+\mathbf B\mathbf u\mathbf u$$ $$\mathbf u=\mathbf u_0 \text{ at } t=0$$ where $\mathbf u$ is the vector of species concentrations, $\mathbf A$ is a matrix specifying the reverse reaction steps, and $\mathbf B$ is a rank 3 tensor specifying the forward reaction steps.
I understand the $\mathbf A\mathbf u$ term, but I am having more trouble understanding the $\mathbf B\mathbf u\mathbf u$ term. I understand that each of the forward reaction terms involve a product of $XY$. So I guess the $\mathbf u\mathbf u$ product is a sort of outer product that gives us a matrix with elements that are all possible products of species (correct me if I am wrong about this). But then what does the $\mathbf B$ tensor even look like, and how is it multiplying the $\mathbf u\mathbf u$ term?
There are different ways to represent a rank $3$ tensor $B$ but in each case, it must have $3$ indices, so this has to be different from $B_i$ with a single index. A $3$-tensor $B = B^i_{jk}$ defines, using Einstein's summation convention, the vector-valued bilinear form
$$Buu := B^i_{jk}u^ju^k$$
that can be plugged into your ODE.
In order to understand the connection with the above formulas, we need to define $u$ in terms of the variables. Assuming $u=(X,Y,Z)$, i.e. the joint collection of all $X_i, Y_i, Z_i$, we can reindex the components of $u$ as
$$u^i = X_i, \quad u^{N+i} = Y_i, \quad u^{2N+i} = Z_i,$$
where $N$ is the dimension of each of $X, Y, Z$, and collect all terms that are quadratic in $u$ into a rank $3$ tensor. For instance, the term $k_i^+X_iY_i$ in the expression for $dX_i/dt$ can be used to write
$$B^i_{j,N+k}u^ju^{N+k} = k_i^+X_iY_i$$
from which we find $B^i_{j,N+k} = k_i^+ \delta_{ij}\delta_{ik}$ using the Kronecker $\delta$ notation.