In the definition of the associated vector bundle $E$ to a principal bundle $\pi:P\rightarrow M$, the equivalence relation is $$(p,v)\sim(pg,g^{-1}v)$$ where $p\in P$, $v\in V$, $g\in G$, Lie group $G$ acts on vector space $V$.
I don't understand the reason/motivation for $g^{-1}$. Why would this cause any problem: $$(p, v)\sim(pg, v)$$
Part of the reason for asking: Say the principal bundle is a frame bundle associated to the vector bundle. Then $g$ is just a basis transformation, which "shouldn't" change a tensorial object $v$. (Meaning it changes the basis of the vector space and the coordinates of $v$ but not $v$ itself, i.e. $v = v^\mu\partial_\nu = v^{\mu'}\partial_{\mu'}$ where $g \partial_\mu = \partial_{\mu'}$).
Let $W$ be an $n$-dimensional real vector space. How does one "write down" a vector $w \in W$? The typical way is to choose a basis $p = (e_1, \ldots, e_n)$ of $W$ and a list of real numbers $\xi_1, \ldots, \xi_n \in \mathbb{R}$, and write $$w = \xi_1e_1 + \cdots + \xi_ne_n.$$ If we think of $p = (e_1, \ldots, e_n)$ as an isomorphism $p \colon \mathbb{R}^n \to W$ --- by way of $p(1,0,\ldots,0) = e_1$, etc. --- and think of $\xi = (\xi_1, \ldots, \xi_n) \in \mathbb{R}^n$ as a vector, then our expression for $w$ is just $$w = p(\xi).$$ But there's some ambiguity here, in the sense that we could have chosen a different basis. In fact, for any invertible matrix $g \in \text{GL}_n(\mathbb{R})$, we could instead take the basis $p \circ g \colon \mathbb{R}^n \to W$ and the list of coefficients $g^{-1}\xi \in \mathbb{R}^n$ to get the same vector: $$(p \circ g)(g^{-1}\xi) = p(\xi).$$ So, the vector $w$ is equally well represented by the pair $(p,\xi) \in P \times \mathbb{R}^n$ as it is by $(p \circ g, g^{-1}\xi) \in P \times \mathbb{R}^n$, where here $P = \{\text{bases of }W\}$. This is the motivation for declaring $(p,\xi) \sim (p \circ g, g^{-1} \xi)$, and the above essentially proves that $$W \cong \frac{P \times \mathbb{R}^n}{\sim}.$$