I want to find the tangent space to the identity for $O(n)$. We can see that $O(n)$ is a smooth manifold, as it is the level-set $F^{-1}(I)$ for $F:M_n(\Bbb R)\to S_n(\Bbb R), F(A)=AA^t$ where $S_n(\Bbb R)$ is the manifold of symmetric matrices.
Then $T_eO(n)=\ker(dF_e)$. We can see that $F= m\circ (id\times (-)^t)\circ \Delta$ where $\Delta(A)=(A,A)$ is the diagonal map, $(-)^t$ is the transpose map and $m$ is the Lie group multiplication of $O(n)$.
Then $dF_e(A)=dm_{(e,e^{t})}\circ d(id\times (-)^t)_{(e,e)}\circ d\Delta_e(A)$ $$=dm_{(e,e')}\circ (id\times d(-)^t)_{(e,e)}(A,A)=dm_{(e,e)}(A,A^t)=A+A^t,$$
so that $T_e O(n) = \{A\in M_n(\Bbb R)\mid A+A^t=0\}$.
Is this correct? I've shown that $dm_g(A,B)=A+B$ for tangent vectors $A,B$ at $g$, and I believe I have shown $d(-)^t(A)=A^t$.
This is correct. In other words $T_e O(n)$ is the sub vector space of skew-symmetric matrix. The fact that $d(-)^t(A)=A^t$ can be written $d_e(-)^t=(-)^t$. This is true because $(-)^t$ is a linear map, hence it is equal to its differential.