Note that $g_1^{-1}x_kg_1 = x_{k+1}$ for all $k$: indeed, $g_1^{-1}x_kg_1 = g_1^{-1}g_1^{-k}x_0g_1^kg_1 = g_1^{-(k+1)} x_0 g_1^{k+1} = x_{k+1}$.
We now proceed by induction on $j$. If $j=1$, then the left hand side is $g_1^{-1}g_2$, and the right hand side is $x_0$, so we have equality.
Assume the asserted equality holds for $j$. Then $\begin{align*} g_1^{-j-1}g_2^{j+1} &= g_1^{-1}\Bigl( g_1^{-j}g_2^j\Bigr)g_2\\ &= g_1^{-1}\Bigl( x_{j-1}x_{j-2}\cdots x_0\Bigr) g_2\\ &= g_1^{-1}x_{j-1}1x_{j-2}1\cdots 1x_01g_2\\ &= g_1^{-1}x_{j-1}(g_1g_1^{-1})x_{j-2}(g_1g_1^{-1})\cdots (g_1g_1^{-1})x_0(g_1g_1^{-1})g_2\\ &= (g_1^{-1}x_{j-1}g_1) (g_1^{-1}x_{j-2}g_1) \cdots (g_1^{-1}x_0g_1) g_1^{-1}g_2\\ &= x_{j}x_{j-1}x_{j-2}\cdots x_1(g_1^{-1}g_2)\\ &= x_jx_{j-1}\cdots x_1x_0, \end{align*}$ as desired.
Note. It doesn't matter that the elements of $G$ are automorphisms, just that you have a group. The fact that $G$ is nonabelian is also irrelevant (though if $G$ is abelian then $x_i = x_0$ for all $i$, and the equation reduces to $(g_1^{-1}g_2)^j = x_0^j$, which is trivially true). And we don't actually need to rename $g_1$ (though it was probably done for some reason in the proof you were reading).