A kind of explanation was given, but I'm missing some fundamental points.
First of let's define the orthogonal Group again as a Manifold M:
$M = O(n) = \{A \in \mathbb{R}^{n\times n}\vert \quad A^T\,A = \mathbb{1}\}$
or
$M = g^{-1}(0)$ with $g = A^T\,A-\mathbb{1}$
Now here comes the tricky party:
A curve $\Gamma$ is being defined where $\Gamma: (-\varepsilon,\epsilon) \to O(n)$ with $\Gamma(0) = \mathbb{1}$
Here what's $\Gamma$ doing? Drawing a curve in orthogonal matrices space?
But how's $\Gamma$ looking?
For instance $\Gamma = \left(\begin{array}{cc}\cos(\varphi) &-\sin(\varphi) \\ \\ \sin(\varphi) & \cos(\varphi)\end{array} \right) \quad \varphi \in (-\epsilon,\epsilon)$ ?
Now there's somehow taken the derivative of $\Gamma^T\,\Gamma = \mathbb{1}$ thus:
$(\Gamma^T\,\Gamma)'(0) = (\mathbb{1})' \quad \Rightarrow \quad (\Gamma^T)'(0)+\Gamma'(0) = 0$
which indicates $\Gamma'(0) \in SO(n) = \{A \in O(n) \vert \quad \det(A) = 1\}$
Furthermore the tangent space $T_pM$ can be included:
$T_pM = \Gamma'(0)$ what means $T_pM \subset SO(n)$
In order to spot this do you really have to use $\Gamma^T\,\Gamma = \mathbb{1}$? Otherswise?
Now additionally the redirection should be shown.
There it is defined: $\Gamma: \mathbb{R}^1 \to O(n)$ where $\Gamma(t) = \exp(t\cdot\Gamma'(0))$
Hence $\Gamma(0) = \mathbb{1}$ and $\Gamma'(0) = \Gamma'(0)$. But what does this mean?
I don't understand the conclusion being made: $SO(n) \subset T_pM$
Logically the tangent space $T_pM$ of $O(n)$ on $p = \mathbb{1}$ being equal to:
$T_pM = SO(n)$ the special orthogonal group. But I don't see why.