The following I know to be true: let $A$ and $B$ be elements of $GL(m,\mathbb{R})$ and let $X \in T_BGl(m, \mathbb{R})$ and let $L_A:Gl(m, \mathbb{R}) \to GL(m, \mathbb{R})$ be the left multiplication map $L_A:B \mapsto AB$. Then the image of $X$ under the associated pushforward map $L_{A*}:T_BGL(m, \mathbb{R}) \to T_{AB}GL(m, \mathbb{R})$ equals $AX$, where the multiplication is understood to be matrix multiplication.
The Clifford algebra $Cl(n, n)$ is isomorphic to the matrix algebra $\mathbb{R}(2^n)$, hence the group of units (invertible elements) $Cl(n,n)^*$ is a Lie group isomorphic to $GL(2^n, \mathbb{R})$. Considering this isomorphism, it seems likely that the above statement for the general linear group holds for the $Cl(n, n)^*$ as well.
More specifically, let $g$ and $h$ be elements of the (Lie) group of units $Cl(n,n)^*$ of the Clifford algebra $Cl(n, n)$. Let $V \in T_hCl(n, n)^*$ and let $L_g:Cl(n, n)^*\to Cl(n, n)^*$ be the left multiplication map in $Cl(n, n)^*$. Is it then true that the image of $V$ under the associated pushforward map $L_{g*}:T_hCl(n, n) \to T_{gh}Cl(n, n)$ equals $gV$, where the multiplication is understood to be the multiplication of the Clifford algebra $Cl(n, n)$?
Bonus question: In fact, all Clifford algebras are isomorphic to matrix algebras over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. Therefore, I would assume the above to hold in the group of units of any Clifford algebra. Correct?