I have seen many different definitions of $G$-principal covers (which I assume are largely equivalent) so in this context, a right $G$-principal cover is a properly discontinuous action of $G$ on a covering $p \colon E \to B$ with $p(xg) = p(x)$ such that $G$ acts transitively on the fibers.
The construction of the fiber transport I am working with is from Tom Dieck, which for a path $w \colon I \to B$ from $b$ to $c$ gives a map $T_p(w) \colon \pi_0(F_b) \to \pi_0(F_c)$, where $F_b$ and $F_c$ are the fibers of $b$ and $c$ respectively. If $x \in F_b$ then there is a lift $v \colon I \to E$ with initial condition $x$, and $T_p(w)$ sends the path component of $x$ to the path component of $v(1) \in F_c$.
In Tom Dieck's algebraic topology book, he claims that it is immediate from the construction that $T_p(w)$ is $G$-equivariant for a right $G$-principal cover, but I'm not really sure why this is true. If $v_x$ is the lift of $w$ at $x$ and $v_{gx}$ the lift at $gx$, it seems to me that this statement is equivalent to the existence of a path from $v_{gx}(1)$ to $v_x(1)g$. However I am not sure how proper discontinuity or transitivity can tell us anything about how compatible an action is with paths.
Using your notation, observe that $v_xg$ - that is, the map $t\mapsto v_x(t)g$ - is a lift of $w$ with $v_xg(0) = v_x(0)g=xg$. The unique path lifting property tells you that $v_xg=v_{xg}$. In particular, the endpoints are the same: $v_{xg}(1)=v_x(1)g$.