As a part of my current homework assignment, I am to derive the first variation of energy identity. Working out the problem with my friends, we came to exactly the same argument as presented in these notes (I have cut out some irrelevant parts from the presentation there, but kept the explanation of terminology and notation; at any rate, I am just using the notes to save myself the work of typing the whole thing).
I understand every step except the second:
$\color{blue}{\bf \large (3)}$ $\nabla$ is a metric connection and $\langle\,\cdot\,,\,\cdot\,\rangle$ is symmetric
$\color{blue}{\bf \large (4)}$ $\nabla$ is torsion-free (this is mentioned above: "where in the fourth equality...")
$\color{blue}{\bf \large (5)}$ $\nabla$ is a metric connection
$\color{blue}{\bf \large (6)}$ fundamental theorem of calculus
What is the justification for step $\color{red}{\bf (2)}$?
We have an ordinary scalar function on $[a,b]\times (-\epsilon,\epsilon)$, which for convenience let's name $h$: $$h(t,s)=\langle \dot{\gamma}_s(t),\dot{\gamma}_s(t) \rangle.$$ We take the partial derivative of $h$ w.r.t. $s$, which is again a scalar function on $[a,b]\times(-\epsilon,\epsilon)$. Now, I'm fine with the equality $$\dot{\gamma}_s(t)=\frac{\partial f}{\partial t},$$ but how exactly does the differentiation $\dfrac{\partial}{\partial s}$ get turned into $\nabla_{\tfrac{\partial f}{\partial s}}$?
Update. Apparently the question is now clear to the OP, but anyway I've decided to summarize the discussion that occurred here for those who are not interested in reading the whole lot.
As the OP commented below @Ted's answer,
After some thought I've realized that we all tried to justify the step (2) which turned out to be wrong and unnecessary. In fact, that comment reveals that merely the domain of the derivative is not correct.
The calculation is still fine despite of the problem with the second step because one simply has to use the fact that $$ \frac{d}{d s} \langle X, Y \rangle = \langle D_s X, Y \rangle + \langle X, D_s Y \rangle $$ where $X$ and $Y$ are vector fields along a curve with parameter $s$, and $D_s$ is the covariant derivative along the curve (see e.g. Lemma 5.2 of J.M.Lee's "Riemannian Manifolds", p.67). Now I think that Proposition 2.2 in M.P. do Carmo's "Riemannian Geometry", p.50, states this fact even in in more suitable form, because in Lee's book we have to return to Lemma 4.9 on p. 57, part (c) that implies that in $X$ is an extendible vector field along the curve then for any extension $\widetilde{X}$ of this field we have $$ D_s X(s) = \nabla_{\frac{\partial f}{\partial s}} \widetilde{X} $$ along the curve.
I keep the text below for the sake of completeness.
I would say that $\frac{\partial}{\partial s}$ at $s=0$ is the same as the variation vector $V$, and for a scalar $h$ we can say that $V\,h = \nabla_V h$
Edit. Let me make my remark more convincing.
The partial derivative of a scalar $h(t,s)$ w.r.t. $s$ is precisely the covariant derivative of $h$ along the curve $p_t (s)$ (for each $t$ fixed). This is just because covariant derivatives of functions (scalars) are the same as the directional derivatives (as @Ted correctly pointed), and they are the same as the Lie derivatives, and the exterior... - they all agree on functions.
By definition, the variation vector field $V$ is the pushforward of the coordinate vector field $\frac{\partial}{\partial s}$, that is $$ V := \mathrm{d}f \left( \frac{\partial}{\partial s} \right) = \frac{\partial f}{\partial s} $$ so we have $$ \frac{\partial}{\partial s} h = \mathrm{D}_s h = \nabla_V h $$
Here is the picture: