$ \def\d{\delta} \def\e{{\mathbf e}} \def\u{{\mathbf u}} \def\x{{\mathbf x}} \def\y{{\mathbf y}} \def\z{{\mathbf z}} \def\<{\langle} \def\>{\rangle} $ If I understand your questions correctly then the answers are 1) Yes 2) No.
Writing $\z=\x+i\y$, we have $f(\z) = \|\x\|^2 - \|\y\|^2+2i\<\x,\y\>$. So $M=f^{-1}(1)$ (resp. $\mathrm T_\z M$) consists of all $\z$ (resp. $\d\z=\d\x+i\d\y$) such that $$ \|\x\|^2 - \|\y\|^2= 1, \qquad \<\x,\y\>=0, \tag1 $$ $$ \llap{(\text{resp.}\qquad}\<\x,\d\x\>-\<\y,\d\y\>=0,\qquad\<\d\x,\y\>+\<\x,\d\y\>=0.) \tag2 $$ Given $\z$ in (1), we can choose an orthonormal basis $(\e_1,\dots,\e_n)$ such that $\x = \|\x\|\e_1$ and $\mathbf y = \|\y\|\e_2$. Then (2) says that $(\d x_1,\d y_1)=\frac{\|\y\|}{\|\x\|}(\d y_2,-\d x_2)$ and hence \begin{align} \omega(\d\z,\d'\z) &= \sum_{i=1}^n(\d x_i\d'y_i-\d'x_i\d y_i)\\ &=\Bigl(1+\tfrac{\|\y\|^2}{\|\x\|^2}\Bigr)(\d x_2\d'y_2-\d'x_2\d y_2) + \sum_{i=3}^n(\d x_i\d'y_i-\d'x_i\d y_i) \end{align} which is obviously nondegenerate.
Specialize to $n=3$ and write $\mathrm{TS}^2=\{w=(\u,\y): \|\u\|=1, \<\u,\y\>=0\}$. Then a diffeomorphism $F:\mathrm{TS}^2\to M$ maps $w$ to $\z=\x+i\y$ where $\x=\u+\y\times\u$ ($\times$ is the vector product). Thus we have \begin{align} (F^*\omega)(\d w,\d'w) &=\<\d\x,\d'\y\>-\<\d'\x,\d\y\>\\ &=\<\d\u+\d\y\times\u+\y\times\d\u,\d'\y\>-\<\d'\u+\d'\y\times\u+\y\times\d'\u,\d\y\>. \end{align} When restricted to the fiber $\mathrm T_\u\mathrm S^2$ (so $\u$ is fixed and $\d\u=0$), this is $$ =\<\d\y\times\u,\d'\y\>-\<\d'\y\times\u,\d\y\>=2\<\u,\d'\y\times\d\y\>. $$ In particular the fibers are not lagrangian for the pulled-back structure, which therefore does not coincide with the canonical (co)tangent bundle structure.