The theorem states:
For any $p \in \mathcal{M}$ there's a neighborhood $W$ of $p$ and a number $\delta > 0$, such that, for every $q \in W$, $\exp_q$ is a diffeomorphism on $B_{\delta}(0) \in T_q M$ and $W \subset \exp_q(B_{\delta}(0))$, that is, $W$ is a normal neighborhood of each of itss points.
The proof is relatively short but I do struggle with a specific bit at the moment.
Proof: Let $\epsilon > 0$, $V$ and $\mathcal{U}$ as defined in Proposition 2.7. Define $F: \mathcal{U} \to \mathcal{M} \times \mathcal{M}$ by $F(q,v) = (q,\exp_q v)$. Recall that $\mathcal{U} \subset TU$, where $U$ is the domain of a system of coordinates $x$ at $p$, with $V \subset x(U)$. Consider around $F(p,0) = (p,p) \in \mathcal{M} \times \mathcal{M}$ the system of coordinates $(U \times U ; (x,x))$. Thus the matrix $dF_{(p,0)}$ is $$ dF_{(p,0)} = \begin{pmatrix} I & I \\ 0 & I \end{pmatrix} $$
I'm not entirely sure how this matrix is actually calculated. By definition $dF_{(p,0)}$ is the Jacobian of the coordinates transformation between $\mathcal{U}$ and $\mathcal{M} \times \mathcal{M}$,
The total number of coordinates is $2n$ if $n$ is the dimension of the manifold $\mathcal{M}$. Since $F = (Id,\exp(\cdot,\cdot))$ The matrix of the Jacobian should have as first $n$ rows $$ \frac{\partial(x_1,\ldots,x_n)}{\partial(x_1,\ldots,x_n)} = I $$ and $$\left. \frac{\partial x^{-1} \circ \exp \circ x}{\partial x_i} \right|_{(p,0)} = \frac{\partial(x_1,\ldots,x_n)}{\partial(x_1,\ldots,x_n)} = I, $$ For the next $n$ rows the coordinates used are in $T_p M$, I'll name them $y_1,\ldots,y_n$. Wrt such variables
$$ \frac{\partial(x_1,\ldots,x_n)}{\partial(y_1,\ldots,y_n)} = 0 $$ While $$ d \exp_{(p,0)} = d (\exp_{p})_0 = \frac{\partial(y_1,\ldots,y_n)}{\partial(y_1,\ldots,y_n)} = I $$ What I'm not sure about (despite I think my logic is correct) is how I'm using the notation to carry out my arguments.
Is my proof correct?
(The proof of the actual theorem continues, but that's the bit I don't get).
Update : The same theorem is stated in "Differential Geometry of Curves and Surfaces" of the same author, however clearly there's the assumption of of regular surfaces instead of manifolds. I'll write down the proof given in that case:
We first show that $d F$ is not singular at $(p,0)$. For that we investigate how $F$ transforms the curves in $\mathcal{U}$ given by $$ \begin{array}{l} t \to (q,tv) \\ t \to (\alpha(t),0) \end{array} $$ where $v \in T_q \mathcal{M}$ and $\alpha(t)$ is a curve in $\mathcal{M}$ with $\alpha(0) = q$. Observe the tangent vectors of such curves at $t=0$ are, respectively $(0,v)$ and $(\alpha'(0),0)$, thus $$ d F_{(q,0)}(0,v) = \left. \frac{d}{dt}(q,\exp_q(vt)) \right|_{t=0} = (0,v) $$ $$ d F_{(q,0)}(\alpha'(0),0) = \left. \frac{d}{dt}(\alpha(t),\exp_{\alpha(t)}(0)) \right|_{t=0} = (\alpha'(0),\alpha'(0)) $$
I've tried to do the same computation by using the definition of differential map between two differentiable manifolds, I get the same result however it seems to me the associated matrix with this transformation $dF$ should be transposed one instead of the one given by the textbook.
However it doesn't really matter because we wanted to prove the non singularity of that matrix but I'd really like to understand where that matrix comes from.
By checking the definition of jacobian matrix, $dF_{(p,0)}$ should equal to $\begin{pmatrix}I&0\\I&I\end{pmatrix}$, consistent with "Differential Geometry of Curves and Surfaces". It should be a typo in "Riemannian Geometry".
For the right-bottom block of $dF_{(p,0)}$, we only need $d(\exp_p)_0=I$. This follows from the proof of Proposition 2.9 of "Riemannian Geometry". Granted that, I think your proof is correct.
Regarding the notation, writing $F=(Id,\exp(\cdot,\cdot))$ is a bit confusing because your "$Id$" corresponds only to the first $n$ arguments.