Given positive integers $m, the implicit function theorem tells you that the zero set of a function $F:\mathbb{R}^n \to \mathbb{R}^m$ is an $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ if the rank of $DF$ is maximal (i.e., equal to $m$) everywhere on that set. In your case $F: \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^2$ is given by $F(a,b) = (\|a\|^2 -1 , a\cdot b)$, and it is easy to check that $DF$ has rank 2 on the set where $F=0$.
The implicit function theorem also gives you charts whose regularity is the same regularity as that of $F$. Since $F$ is $C^\infty$, your charts are $C^\infty$, too. Since the projection map in the ambient space $\mathbb{R}^3 \times \mathbb{R}^3$ is $C^\infty$, you get the same regularity on your manifold.
If your definition of manifold includes connectedness, it is pretty straightforward to check, too. Any point $(a,b)$ on $TS^2$ can be connected to $(a,0)$ via $t \mapsto (a,tb)$, and the rest follows from path-connectedness of $S^2$.