Derivative of the symplectomorphism evaluated at a point of the zero section of the cotangent bundle

Question

It might be an easy question, possibly not worth posting here. In the proof of the Lagrangian neighborhood theorem, the authors have written the expression for the derivative of a symplectomorphism at a point of the zero section. I don't understand how one arrives at this expression.

Let $L$ be a Lagrangian submanifold of the symplectic manifold $(M,\omega)$. Choose an almost complex structure $J\colon TM\to TM$ compatible with $\omega$. Thus $g_J:=\omega(-,J\cdot)$ is a Riemannian metric on $M$.

Define $\phi\colon T^*L\to M$ as follows $\phi(q, v^*)=\exp_q\left(J_q \Phi_q(v^*)\right)$, where $g_J\left(\Phi_q(v^*),v\right)=v^*(v)$.

Question 1: I think $\phi$ can't be defined on whole $T^*L$, shouldn't the domain of definition of $\phi$ is an open subset of the zero section $L\subset T^*L$?

Question 2: Why $d\phi_{(q,0)}(v_0,v_1^*)=v_0+J_q\Phi_q(v_1^*)$, where $(v_0,v_1^*)\in T_qL\oplus T^*_qL=T_{(q,0)}T^*L$?

Note that $J(TL)$ is the normal bundle $NL$ of $L$ in $M$ w.r.t. $g_J$.

Reference: Introduction to Symplectic Topology by Dusa McDuff, Dietmar Salamon; see page 121.

Answer 1

On Question 1, $\phi$ can be defined on the whole of $T^*L$ so long as the exponential map can be defined on the normal bundle $NL$, which will be the case for example if $M$ is geodesically complete. It is not generally 1-1 on $T^*L$, but is on a neighbourhood of the zero section $L$ - is this possibly what you were thinking?

On Question 2, $\phi(q,0) = q \implies d\phi_{(q,0)}(v_0,0) = v_0$ while $$d\phi_{(q,0)}(0,v_1^*) =\frac{d}{dt}\Big\vert_{t=0}\phi(q,tv_1^*) = \frac{d}{dt}\Big\vert_{t=0}\exp_q(tJ_q\Phi_q(v_1^*)) = J_q\Phi_q(v_1^*)$$ by definition of the exponential map.