Related to this question, which has a reference to the statement of the lemma in the subject of this question.
I'm trying to work out by myself the details of equation (3) of the book.
$$ \left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right\rangle(1,0) = \left\langle d (\exp_p)_v(w_N), d (\exp_p)_v(v) \right\rangle $$
Although it has been answered already in one of the linked questions I've been trying to add more details just as an exercise. Also mostly because I want to make sure I understand the notation used by author.
For sake of making things a bit shorter I'll use the notation
$$ \begin{array}{l} \partial_s f = \frac{\partial f}{\partial s} = df_{(t,s)} \left( \partial_s\right)\\ \partial_t f = \frac{\partial f}{\partial t} = df_{(t,s)} \left( \partial_t\right) \end{array} $$
Since
$$ f : (t,s) \to \exp_p(tv(s)) $$
where $v(s)$ is a curve in $T_p \mathcal{M}$ such that $v(0) = v$ and $v'(0) = w_N$ we have
$$ \partial_s f = df_{(t,s)} \left( \partial_s \right) = d (\exp_p)_{(t,s)} \left( \partial_s \right) $$ $$ \partial_t f = df_{(t,s)} \left( \partial_s \right) = d (\exp_p)_{(t,s)} \left( \partial_t \right) $$
Therefore
$$ \left\langle \partial_s f , \partial_t f\right\rangle(1,0) = \left\langle d (\exp_p)_{(t,s)} \left( \partial_s \right), d (\exp_p)_{(t,s)} \left( \partial_t \right) \right\rangle(1,0) \;\;\; (1) $$
Is my derivation of (1) correct? Assuming it is...
I think the difficult bit is to compute $d (\exp_p)_{(t,s)}$, my attempt is the following
$$ d (\exp_p)_{(t,s)} = d (\exp_p)_{tv(s))} d (tv(s))_{(t,s)} \;\;\;\; (2) $$
I'll know compute $d (tv(s))_{(t,s)}$. Since $v : s \to T_p \mathcal{M}$ it means it can be written as
$$ tv(s) = \sum_{i=1}^n t v_i(s) \partial_{x_i} = t v_i (s) \partial_{x_i} $$ Since the differential of the map is the Jacobian of the coordinates transformation we have
$$ d(tv(s))_{(t,s)} = \begin{pmatrix} v_1(s) & tv_1'(s) \\ \vdots & \vdots \\ v_n(s) & tv_n'(s) \end{pmatrix} $$
If I substitute $(t,s) = (1,0)$ I get
$$ d ( \exp_p )_{tv(s)} = d ( \exp_p )_{v(0)} = d ( \exp_p )_{v} $$ $$ d ( tv(s))_{(t,s)} = \begin{pmatrix} v_1(s) & tv_1'(s) \\ \vdots & \vdots \\ v_n(s) & tv_n'(s) \end{pmatrix} = \begin{pmatrix} v_1(0) & v_1'(0) \\ \vdots & \vdots \\ v_n(0) & v_n'(0) \end{pmatrix} $$
Observe in the matrix above the first column represents the coordinates of $v$ while the second column represents the coordiantes of $w_N$. Plugging $\partial_s$ would provide me the second column, while $\partial_t$ would provide me the first, plugging everything together I get the equation (3) of the book.
There few details I'm omitting but I think the idea is clear, I also hope I haven't done any mistakes with rows and columns in the computation of the differential map.