This is a technical lemma about the existence of a smooth curve between any two points on a sphere.
Lemma: Let $\mathcal{S}=\{ x:||x||=r\}$, where $r>0$, and let $t_1,t_2\in \mathcal{S}$. Then there is a differentiable mapping $c:(0, 2\pi) \mapsto \mathcal{S}$ that passes through $t_1$ and $t_2$; that is, there exist $\alpha_1,\alpha_2 \in (0, 2\pi)$ such that $c(\alpha_1)=t_1$ and $c(\alpha_2)=t_2$.
Proof: Let $\mathcal{H}$ be a hyperplane that passes through $t_1,t_2,0$ in $\mathbb R^p$. Then $\mathcal{C}=\mathcal{H} \cap \mathcal{S}$ is a circle centered at the origin with radius $r$. Pick an arbitrary point $t_3$ in $\mathcal{C}$ that is not $t_1$ and $t_2$, and let $t_4$ be the point on $\mathcal{C}$ obtained by rotating $t_3$ counterclockwise 90 degree along $\mathcal{C}$. Then any point $t$ in $\mathcal{C}$ that is not $t_3$ can be written as $t=\cos(\alpha)t_3+\sin(\alpha)t_4$ for some $\alpha \in (0, 2\pi)$. By construction, $\{ c(\alpha): \alpha \in (0, 2\pi) \}$ is a differentiable curve that passes through $t_1$ and $t_2$.
Question 1: In the proof, it lacks the description of the function $c(\alpha)$. It only says that $t=\cos(\alpha)t_3+\sin(\alpha)t_4$ for some $\alpha \in (0, 2\pi)$. But what is the function $c(\alpha)$?
Question 2: For any point $t \in C$, why we can guarantee that t can be expressed as $\cos(\alpha)t_3+\sin(\alpha)t_4$? I know $t_3$ and $t_4$ are orthogonal. I also know $H$ is a 2-dimensional plane and $C$ is a circle centered at the origin with radius r. I guess it's due to Fourier basis?
Question 3: I don't understand that "By construction, $\{ c(\alpha): \alpha \in (0, 2\pi) \}$ is a differentiable curve that passes through $t_1$ and $t_2$.".
As for question 1, presumably $c(\alpha) = t_3\cos(\alpha) + t_4\sin(\alpha)$ to illustrate the conclusion of the lemma.
For question 2, you don't need any Fourier analysis. It's just that once you identify that $t_3$ and $t_4$ are orthogonal vectors in a vector space and they have the same modulus, then in the plane spanned by $t_3$ and $t_4$, the curve $c(\alpha)$ can be identified with a circle of radius $|t_3| = |t_4|$, the same way that $e_1\cos(\theta) + e_2\sin(\theta)$ is a unit circle in the plane, and $e_1,e_2$ is the standard orthonormal basis of the plane.
For question 3, the curve $c(\alpha)$ is differentiable because $\cos(\alpha)$ and $\sin(\alpha)$ are differentiable.