$
\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}
$$
\def\C{{\mathbf C}}
\def\d{\delta}
\def\e{{\mathbf e}}
\def\r{{\mathbf r}}
\def\u{{\mathbf u}}
\def\x{{\mathbf x}}
\def\y{{\mathbf y}}
\def\z{{\mathbf z}}
\def\<{\langle}
\def\>{\rangle}
$
IfIf I understand Now that I understand your questions correctly then, the answers are 1) Yes 2) NoNo Yes.
Writing $\z=\x+i\y$, we have $f(\z) = \|\x\|^2 - \|\y\|^2+2i\<\x,\y\>$. So $M=f^{-1}(1)$ (resp. $\mathrm T_\z M$) consists of all $\z$ (resp. $\d\z=\d\x+i\d\y$) such that $$ \|\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 $$ Given $\z$ in (1), we can choose an orthonormal basis $(\e_1,\dots,\e_n)$ such that $\x = \|\x\|\e_1$ and $\mathbf y = \|\y\|\e_2$. Then (2) says that $(\d x_1,\d y_1)=\frac{\|\y\|}{\|\x\|}(\d y_2,-\d x_2)$ and hence \begin{align} \omega(\d\z,\d'\z) &= \sum_{i=1}^n(\d x_i\d'y_i-\d'x_i\d y_i)\\ &=\Bigl(1+\tfrac{\|\y\|^2}{\|\x\|^2}\Bigr)(\d x_2\d'y_2-\d'x_2\d y_2) + \sum_{i=3}^n(\d x_i\d'y_i-\d'x_i\d y_i) \end{align} which is obviously nondegenerate. (Or as Tim Perutz says, just note that $M$ is a complex submanifold.)
Specialize to $n=3$ and write $\mathrm{TS}^2=\{w=(\u,\y): \|\u\|=1, \<\u,\y\>=0\}$. Then a diffeomorphismConsider $F:\mathrm{TS}^2\to M$ maps(after Seidel's thesis pointed out by Daniel Pomerleano) the diffeo $w$ to$\Phi:M\to\mathrm{TS}^{n-1}$ sending $\z=\x+i\y$ whereto the pair $\x=\u+\y\times\u$$(\u,\r)=(\frac{\x}{\|\x\|}, \|\x\|\y)$. Writing ($\times$ is$\sigma$ for the vector product). Thuscanonical 2-form $d\<\u,d\r\>$ on $\mathrm{TS}^{n-1}$, 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}\begin{align} \Phi^*\sigma &= d(\Phi^*\<\u,d\r\>)=d(\<\tfrac{\x}{\|\x\|},d(\|\x\|\y)\>)\\ &=d\<\x,d\y\> + \<\x,\y\>\tfrac{d\|\x\|}{\|\x\|} \end{align} When restricted to the fiberand this equals $\mathrm T_\u\mathrm S^2$$\omega$ since the second term vanishes by (so1). Thus $\u$$\Phi$ is fixeda symplectomorphism.
My initial (stupid and wrong) answer came from my thinking of another similar question. Namely $\d\u=0$$\mathrm{TS}^2$, thought of as the manifold of oriented affine lines in $\mathbf R^3$ (identify $(\u,\r)$ with the line $\r+\mathbf R\u$ with orientation $+\u$), this iscarries a 2-parameter family of Euclid-invariant symplectic structures given by $$ =\<\d\y\times\u,\d'\y\>-\<\d'\y\times\u,\d\y\>=2\<\u,\d'\y\times\d\y\>. $$$$ \omega_{k,s} = d\<k\u,d\r\> - s\Omega,\qquad k>0, s\in \mathbf R, \tag3 $$ In particularwhere the fibers are not lagrangian forfirst term is the pulledcotangent bundle 2-back structureform, which therefore does not coincide withand the canonicalsecond (co"magnetic")tangent bundle structure term is $-s$ times the area 2-form of the base $\mathrm S^2$.
On the other hand, identifying $\mathfrak{so}(3,\C)=\C^3$ with Lie bracket the vector product, the nonzero levels of your quadratic form $\<\z,\z\>=\sum_{i=1}^3 z_i^2$ are precisely the semisimple (co)adjoint orbits of $\mathrm{SO}(3,\C)$. As such they carry a (Kirillov-Kostant-Souriau) 2-form $\omega_{\mathrm{KKS}}(\d\z,\d'\z)=\mathrm{Re}\bigl(\frac1{s^2}\<\z,\d'\z\times\d\z\>\bigr)$ where $\<\z,\z\>=s^2$ is the equation of our orbit $\mathcal O$. So we can again ask if this is symplectomorphic with $\mathrm{TS}^2$ equipped with one of the forms (3). At least when $s^2$ is real positive, the answer is yes and the diffeomorphism $F:\mathrm{TS}^2\to\mathcal O$ I was considering (with $k=s=1$), $$ F(\u,\r) = k\r\times\u + s\u + ik\r $$ is symplectic: $F^*\omega_{\mathrm{KKS}}=\omega_{k,s}$. Being familiar with this diffeomorphism, I then made the mistake of thinking it must be "the one" of which you were asking if it was symplectic relative to your (different) 2-forms. And it isn't.