Which matrices can be realized as second derivatives of orthogonal paths?

I believe that the construction you're looking for is called a Dyson Series (Wikipedia). In detail, suppose we are given $B,C \in \mathrm{Skew}(n)$ and we want to construct a $\gamma: (-\varepsilon,\varepsilon) \to \mathrm{SO}(n)$ such that $\gamma(0)=1$ and $\ddot{\gamma}(0)=B^2+C$. I claim that \begin{align*} \gamma(t) & :=\sum_{n=0}^{\infty} \left[\int_0^t \int_0^{t_0} \cdots \int_0^{t_{n-1}} \left(\prod_{k=0}^n (B+t_{n-k} C) \right) \mathrm{d} t_n \cdots \mathrm{d} t_0\right] \\ & = 1 + \int_0^t (B+t_0C) \mathrm{d}t_0 + \int_0^t \int_0^{t_0} (B+t_1C)(B+t_0C) \mathrm{d} t_1 \mathrm{d}t_0 + \\ & \hspace{1cm}\int_0^t \int_0^{t_0} \int_0^{t_1} (B+t_2C)(B+t_1C)(B+t_0C) \mathrm{d}t_2 \mathrm{d}t_1 \mathrm{d}t_0 + \cdots \end{align*} is a well-defined solution to the problem. Indeed, if we let $$m:=\max_{s \in [0,t]} \lVert B+sC \rVert_{L^2},$$ then we have that $$\lVert \gamma(t) \rVert_{L^2} \leq e^m,$$ so that $\gamma$ is defined by a convergent sequence. Moreover, we can compute that $\dot{\gamma}(t)=\gamma(t)(B+tC)$, evincing both that the image of $\gamma$ (which a priori lies in the space of $n \times n$ matrices) in fact lies in $\mathrm{SO}(n)$ and also that $\ddot{\gamma}(0)=B^2+C$.

Yes. Given skew-symmetric matrices $B$ and $C,$ define $\alpha(t)=\exp(Bt+\tfrac12 Ct^2).$ Then \begin{align} \alpha(t) &=I+(Bt+\tfrac12 Ct^2)+\tfrac12 (Bt+\tfrac12 Ct^2)^2+O(t^3)\\ &=I+Bt+\tfrac12 (B^2+C)t^2+O(t^3) \end{align} as $t\to 0.$ This shows that $\ddot \alpha(0)=B^2+C.$