The following is an excerpt from a proof that the orthogonal group $\operatorname{O}(n)$ is a manifold
Let $f: \mathcal{M} \simeq \mathbb{R}^{n^2}\to \mathcal{S} \simeq \mathbb{R}^{\frac{n(n+1)}{2}}$ where $A \mapsto AA^T$. This function takes a Matrix $A$ and maps it to the symmetric Matrix $AA^T$. We have $f^{-1}(\{I\}) = \operatorname{O}(n)$. We will show that $I$ is a regular value of $f$. First compute the derivative $df(A)(H) = \lim_{h \to 0} \frac{f(A + hH) - f(A)}{h} = AH^T + HA^T$.
Why are we computing the directional derivative of $f$ along the vector $H$ instead of computing the Jacobian matrix? To show that $I$ is a regular value we need to go on and prove that the differential $df(A)$ is surjective for every matrix $A \in \operatorname{O}(n)$. We can obtain this result by letting $M = AS/2$ for an arbitrary symmetric $S \in \mathcal{S}$ and computing $df(A)(M)$. This part is pretty straightforward, but why does it work? Does the differential coincide with the directional derivative in some way?
You are saying how $df_A$ acts on vectors. You are estabilishing the derivative, not just the directional derivative. Computing the Jacobian matrix would be cumbersome and, in some sense, useless. So we know how the derivative itself acts: it takes $H \mapsto AH^T+HA^T$.
Perhaps a better way to expose this would be to compute explicitly as follows:
$$f(X+H)=(A+H)(A+H)^T=(A+H)(A^T+H^T)=AA^T+AH^T+HA^T+HH^T,$$
where the last term is $o(H)$ and the "middle" term is clearly linear.
Note, however, that in the way the text does it, there are some hidden details. The text really just computed directional derivatives. Computing the "directional derivatives in the direction of every vector" gives the derivative if the function is differentiable. Therefore, he must justify that in some way, otherwise there are gaps.