ad 1.): No, the answer is that $\Omega^{1,0} CP^N$ corresponds to $N+1$. Proof sketch:
As complex vector bundles, $TCP^N \oplus C= C^{N+1} \otimes L$, $L$ the tautological line bundle (the one with holomorphic sections). Thus $L^{\otimes (N+1)} = \Lambda^{N+1} (TCP^N \oplus C) = \Lambda^N TCP^N$. Dualizing gives the answer. For a more polished reasoning and more details, consult Griffith-Harris.
ad 2.) I am not sure whether your interpretation of $S_k$ is correct; Friedrich writes (bottom of page 77) that $S_0$ is the
determinant line bundle and $S_N$ is trivial.
Here is how I would figure out the answer, switching perspective on $spin^c$ structures slightly (I do not have a handy reference for the
following, I followed hints I found in Weyl's "The classical groups").
Let $V \to X$ be an oriented vector bundle of rank $2n$. A spin$^c$ structure is a bundle of complex $Cl(V)$-modules that is
(fibrewise) irreducible. Given a line bundle $L$ on $X$, you can tensor a $CL(V)$-module with $L$ and obtain a new module. This gives an action of the group
of line bundles on the set of spin^c-structures.
Note that since $2n$ is even, there is a unique irreducible representation of $Cl(R^{2n})$. Moreover, $Cl_{\mathbb{C}}(R^{2n})$ is
the matrix algebra $\mathbb{C}(2^n)$, so the irrep has to be $2^n$-dimensional. However, this module is determined only up to isomorphism, which is responsible
for the fact that not any vector bundle is spin^c.
Now assume that a complex structure has been given to $V$. Then
the exterior algebra $\Lambda_{\bC} V$, together with the Clifford action defined by the formula $c(v)=v \wedge - \iota_v$ is
a Clifford module. It has to be irreducible for dimension reasons!
This shows that any complex vector bundle is spin^c and gives an explicit desciption of the spin^c structures. Note that the complex
structure was used here essentially, to write the exterior algebra $\lambda_{\bR} V$ (which is a Clifford module, but not irreducible) as
a tensor product
of $\Lambda_{\mathbb{C}}(V)$ with its conjugate.
The result is pretty close to what you wrote and identical if you stick to Friedrichs notation.
Now the clifford algebra is graded
$Cl(V)_{\mathbb{C}}=Cl_{\mathbb{C}}(V)^0 _{\mathbb{C}}(V)^1$ and if you trace back the definition of the action, you see that $Cl_{\mathbb{C}}(V)^0$
preserves the even/odd-decomposition of the exterior algebra. This gives the grading of the spinor bundle, more or less as you suspected.
As I said before, spin^c structures can be twisted by line bundles (tensor products). The only line bundle available on a complex manifold is
the determinant bundle of the tangent bundle (and its powers). By choosing different powers, you can switch between the anticanonical/canonical spin^c structures.
OK, I am making an assumption: I can re-interpret the problem (using the musical isomorphism) as $V$ being the diagonal embedding of $TM$ inside $TM\oplus TM$ and studying spin structures on them.
Actually, it turns out (see comments) that this "re-interpretation" is slightly different from the original construction. But the main bulk still goes through:
I will restrict to $\dim M=3$, in which case our (closed oriented) manifold is always spinnable. A spin structure $\mathfrak{s}$ on $M$ induces a canonical spin structure $\mathfrak{S}_0=\mathfrak{s}\oplus\mathfrak{s}$ on $TM\oplus TM$, and this is actually independent of the choice of $\mathfrak{s}$ (these appear in the notion of a 2-framing on 3-manifolds, which Atiyah and Witten have used for some of their QFT studies). As a result, the "restriction" $\mathfrak{S}_0|_V$ on $M$ is ill-defined.
[proof of claim of canonical spin structure (learned from conversation with Rob Kirby): the spin structure fixes a trivialization over the 1-skeleton, and over circles there are two trivializations, so changing a trivialization of $\mathfrak{s}$ is doubled in $\mathfrak{s}\oplus\mathfrak{s}$ which modulo-2 is no change.]
This also implies that the "restriction" to each $TM$-summand is ill-defined. (What I know that works: the collar-neighborhood theorem does allow an induced spin structure on $T(\partial X$) from a spin structure on $TX$ thanks to the splitting $TX|_\partial=T(\partial X)\oplus\underline{\mathbb{R}}$ near the boundary.)
Best Answer
First let me point out that the bundle $\Lambda^2_+ T^*M$ makes sense only when $\dim M=4$. Maybe you should add this assumption.
As for your question, I think that you can find the answer in Example 1.3.3 page 30 of these notes.