What does algebra $A \rtimes G$ stand for? Let $k$ be an algebraically closed field, $A$ a finite dimensional (unital associative) $k$-algebra of finite dimension, and $G$ a torus over $k$ acting on $A$.

What does $k$-algebra $A \rtimes G$ stand for? More specifically how to define addition, multiplication, and scaler multiplication?

The only usage of $\rtimes$ that I know is that of semidirect product and trivial extension algebra. But none of them make sense here. I came across this in the proposition 5.1 in this paper (alternative link); the author says the statement is classic and doesn't explain what he meant by the symbol. Could you help me?
In a semidirect product $A\rtimes B$, the group $B$ acts on $A$. That is, there is a homomorphism from $B$ into $\operatorname{Aut}(A)$.

Here, you have $G$ acting on $A$, but in a different context. So maybe that is motivating using the notation $A\rtimes G$.

With groups, $A\rtimes B$ is the collection of pairs $(a,b)$, and there is a new group operation $(a_1,b_1)(a_2,b_2)=(a_1a_2^{b_1},b_1b_2)$ where $a_2^{b_1}$ is the image of $a_2$ under the action from $b_1$.

In OP's setting $A\rtimes G$ is the collection of pairs $(a,g)$, with $k(a,g)=(ka,g)=(a,kg)$. You again can have $(a_1,g_1)(a_2,g_2)=(a_1a_2^{g_1},g_1g_2)$, making $A\rtimes G$ an algebraic group. I'm not sure you can get an algebra out of this though, since the $G$ has no addition.
In a semidirect product $A\rtimes B$, the group $B$ acts on $A$. That is, there is a homomorphism from $B$ into $\operatorname{Aut}(A)$.

Here, you have $G$ acting on $A$, but in a different context. So maybe that is motivating using the notation $A\rtimes G$.

With groups, $A\rtimes B$ is the collection of pairs $(a,b)$, and there is a new group operation $(a_1,b_1)(a_2,b_2)=(a_1a_2^{b_1},b_1b_2)$ where $a_2^{b_1}$ is the image of $a_2$ under the action from $b_1$.

In OP's setting $A\rtimes G$ is the collection of pairs $(a,g)$, with $k(a,g)=(ka,g)=(a,kg)$. You again can have $(a_1,g_1)(a_2,g_2)=(a_1a_2^{g_1},g_1g_2)$, making $A\rtimes G$ an algebraic group. I'm not sure you can get an algebra out of this though, since the $G$ has no addition.