Let $M$ a manifold and $T_pM$ it's tangent plan at $p$. We defined \begin{align*} \exp_p:U_p\subset \Omega _p&\longrightarrow M\\ V&\longmapsto \gamma _V(1) \end{align*} where $\gamma _V:I_V\longrightarrow M$ is a geodesic s.t.
1) $\gamma _V(0)=p$,
2) $\dot\gamma_V(0)=V$,
3) $I_V\subset \mathbb R$ is an interval that contain $0$ and $1$.
and $\gamma _V$ is maximal for 1) and 2). And $\Omega _p=\{V\in T_pM\mid 1\in I_V\}.$
Question 1 : First of all,what does mean "$\gamma _V$ is maximal for 1) and 2)" ? What does it mean ?
Question 2 : Then, I don't understand why $$\mathrm d _0\exp_p=id.$$
To me if $f\in \mathcal C^\infty (M)$, then $$\mathrm d _0 f(V)=V(f)$$ where $V\in T_pM$. But how it work here ?
Q1) It's says that if there is a geodesic $\tilde\gamma :I_W\longrightarrow M$ s.t. $\gamma (0)=p$ and $\dot\gamma (0)=V$, then $\gamma_V|_{I_W}=\tilde \gamma $.
Q2) You should have proved somewhere that $\Omega _p$ is star shaped (in $0$), and that $\gamma _V(t)=\exp(tV)$ (which is well defined by the fact that $\Omega _p$ is star shaped). Now, let $V\in \Omega _p$. $$(\mathrm d _0\exp_p)(V):=V\exp_p\underset{(*)}{=}\left.\frac{\mathrm d }{\mathrm d t}\right|_{t=0}\exp(tV)=\left.\frac{\mathrm d }{\mathrm d t}\right|_{t=0}\gamma _V(t)=\dot\gamma (0)=V.$$ Therefore, it's the identity.
Just a justification for $(*)$: It comes from the fact that $\Omega _p$ is star-shaped (in $0$). And thus, every vector of $\Omega _p$ is of the form $0+Vt=Vt$.