The usual action of $fg$ on $u⊗v$ , where $f,g$ are elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$ of $G$, is given by $fg(u⊗v)=fgu⊗v+fu⊗gv+gu⊗fv+u⊗fgv$, using the comultiplication, right? 
How to state this fact for $V^{\otimes n}$, i.e. $fg$ acting on $u⊗v$, where $u=\otimes\_{i=1}^{n-k} u\_i$ and $v=\otimes\_{i=1}^k v\_i$, for each $k=1,...,n-1$ ? 
Thanks,



How to define the action of $U(G)$ in this situation?
There is nothing deep here. The coproduct $\Delta : U(\mathfrak g) \to U(\mathfrak g)\otimes U(\mathfrak g)$ simply implements Leibniz's product rule: if $v\_1\in V\_1$ and $v\_2\in V\_2$, then $x\in \mathfrak g$ acts on $V\_1\otimes V\_2$ by $x: v\_1\otimes v\_2 \otimes (xv\_1)\otimes v\_2 + v\_1\otimes (xv\_2)$. Extending this, on a large tensor product we have:
$$ x(v\_1 \otimes v\_2 \otimes \cdots \otimes v\_n) = xv\_1 \otimes v\_2 \otimes \cdots \otimes v\_n + v\_1 \otimes xv\_2 \otimes \cdots \otimes v\_n + \cdots + v\_1 \otimes v\_2 \otimes \cdots \otimes xv\_n$$
If $x,y\in \mathfrak g$, then $xy \in U(\mathfrak g)$ acts via the composition $x\circ y$. For example,
$$ xy \left( \bigotimes\_{k=1}^n v\_k\right) = \sum\_{k=1}^n (xy\text{ acts on }v\_k) + \sum\_{j\neq k}(x\text{ acts on }v\_j\text{ and }y\text{ acts on }v\_k)$$


Note that in general, for $f\in U(\mathfrak g)$, we do not have $f(u\otimes v) = fu \otimes v + u\otimes fv$. For example, the constant $1$ is an element of $U(\mathfrak g)$, and $1(u\otimes v) = u\otimes v \neq 2u\otimes v = 1u\otimes v + u\otimes 1v$. More generally, not all differential operators are derivations: certainly you do not expect $\frac{\partial^2}{\partial x^2}$ to satisfy a Leibniz rule!


