Lie bracket of vector fields and differential of diffeomorphism in its definition

To make the connection to the Lie derivative, let $t \mapsto \Phi^V_t$ be the 1-parameter group of diffeomorphisms (or flow) generated by the vector field $V$. The differential $d\Phi^V_t$ of each diffeomorphism maps the vector field Y to a new vector field $\mathrm{d}\Phi^V_{t}(Y)$. To pull-back the vector field one applies the differential of the inverse, $d ((\Phi^V_{t})^{-1})= d \Phi_{-t}^V$.

So what is $(\Phi^V_{t})^{-1}$ really saying? According to my knowledge, $\Phi^V_{t}$ is basically a point on flow curve (induced by vector field $V$ of some manifold $M$) given by input $x$ on manifold $M$ and some "time" $t$. So what would inverse map to?

$\Phi_t^V:M\to M$ is invertible and $(\Phi_t^V)^{-1}=\Phi_{-t}^V$. Indeed, if you flow $t$ units of time and then you flow $-t$ units of time, i.e. you flow back $t$ units of time, you end at the same point.
Formally, this is a direct consequence of the "group law" for flows: $\Phi_{s+t}=\Phi_s \circ \Phi_t$, and the fact that $\Phi_0=\operatorname{id}$.
So, applying $(\Phi_t^V)^{-1}=\Phi_{-t}^V$ to a point of $M$ you flow back $t$ units of time.