I've been struggling to solve the following:
Show that $TO(n)$ is diffeomorphic to $O(n; \mathbb{R})\times M(n; \mathbb{R})$.
Here $TO(n; \mathbb{R})$ represents the tangent bundle of $O(n)=\{A\in M(n) ; AA^{t}=Id_{n}\}$.
EDIT: as I said in the comments, I thought that maybe it is $\rm Antsym$ set instead of $M(n)$, because I already verified that for $A\in {\rm Antsym}$ and $B\in O(n)$, $BA\in T_{B}O(n)$. But I wrote down the assertion that I had. Thanks.
Indeed there is a mistake in the statement of the problem, and what you have there is a general phenomenon about Lie groups. It is always true that $TG \cong G \times \mathfrak{g}$, where $\mathfrak{g} = T_eG$ is the Lie algebra of $G$. The diffeomorphism is just $$TG \in v_g \mapsto (g, {\rm d}(L_{g^{-1}})_g(v)) \in G \times \mathfrak{g},$$where $L_g\colon G \to G$ is left-translation by $g$. The inverse is given by $$G\times \mathfrak{g} \ni (g, w) \mapsto {\rm d}(L_g)_e(w) \in TG.$$In your case, we have that $\mathfrak{o}(n) \doteq T_{{\rm Id}_n}{\rm O}(n) = \{X \in {\rm Mat}(n) \mid X+X^\top = 0\}$ and so $$T{\rm O}(n) \cong {\rm O}(n)\times \mathfrak{o}(n).$$The equation describing $\mathfrak{o}(n)$ is deduced as follows: take a curve $A(t)$ of orthogonal matrices with $A(0) = {\rm Id}_n$ and $A'(0) = X$. Differentiate both sides of $A(t)A(t)^\top = {\rm Id}_n$ at $t=0$ to get $X + X^\top = 0$.