Let $g, g'$ be Lie algebras. Let $V$ (resp. $V'$) be a $g$-module (resp. $g'$-module). Do we have a $g \otimes g'$-action on $V \otimes V'$? In particular, when $g=g'$ and $V = V'$, do we have a $g \otimes g$-action on $V \otimes V$? Any help will be greatly appreciated!
Edit: or do we have a $g \oplus g'$-action on $V \otimes V'$? How is the map: $$ (g \oplus g') \times (V \otimes V') \to V \otimes V' $$ defined explicitly?
The tensor product of (the underlying vector spaces of) two Lie algebras does not have a natural Lie algebra structure. You have a $\mathfrak{g} \oplus \mathfrak{g}'$-action. It's defined by extending
$$(X \oplus X')(v \otimes v') = Xv \otimes v' + v \otimes X' v'$$
linearly, where $X, X' \in \mathfrak{g}, \mathfrak{g}'$ and $v, v' \in V, V'$.
In general, if $A, B$ are two algebras and $M, N$ are two modules over $A, B$ respectively, then $M \otimes N$ naturally has the structure of an $A \otimes B$-module. To specialize this to the case of Lie algebras you should let $A, B$ be universal enveloping algebras, then use the identity
$$U(\mathfrak{g} \oplus \mathfrak{g}') \cong U(\mathfrak{g}) \otimes U(\mathfrak{g}').$$
(This identity suggests that one should think of the passage to the universal enveloping algebra as "exponentiating" a Lie algebra, since it converts direct sums to tensor products.)