Other posters explain some of the topology of spin structures. Here's a differential-geometric answer relevant to Dirac operators. The exercise you have set yourself, of understanding Dirac operators on the 2-torus, is a good one. Rather than trying to do it for you, I'll instead discuss the 3-torus (cf. Kronheimer-Mrowka, "Monopoles and 3-manifolds").
So: a spin-structure on a Riemannian 3-manifold $Y$ can be understood in the following workmanlike way: we give a rank 2 hermitian vector bundle $S \to Y$ (the spinor bundle); a unitary trivialization of $\Lambda^2 S$; and a Clifford multiplication map $\rho \colon TY\to \mathfrak{su}(S)$, such that at each $y\in Y$ there is some oriented orthonormal basis $(e_1,e_2,e_3)$ for $T_y Y$ such that $\rho(e_i)$ is the $i$th Pauli matrix $\sigma_i$. More invariantly, one can instead say that $\rho$ is an isometry (with respect to the inner product $(a,b)=tr(a^\ast b)/2$) and satisfies the orientation condition $\rho(e_1)\rho(e_2)\rho(e_3)=1$.
If we have two spin-structures, with spinor bundles $S$ and $S'$, we can look at the sub-bundle of $\mathrm{SU}(S,S')$ consisting of those fibrewise special isometries that intertwine the Clifford multiplication maps. This bundle has fibre $\{ \pm 1 \}$: it is a 2-fold covering of $Y$. As such it is classified by a class in $H^1(Y;\mathbb{Z}/2)$, whose non-vanishing is clearly the only obstruction to isomorphism of the two spin-structures. Conversely, by tensoring everything by real orthogonal line bundles (work out what this means concretely!), you can construct all spin structures, up to isomorphism, from a chosen one.
On flat $T^3$, all the data can be taken translation invariant. The Dirac operator is then $D = \sum_i{\sigma_i\partial_i}$. Tensoring with an orthogonal line bundle $\lambda$ (constructed, if you will, from a character $\pi_1(T^3)\to O(1)$) the formula becomes $D_\lambda =D \otimes 1_\lambda$.
In 2 dimensions, the story will be similar; the new feature is that the spinor bundle splits into two line bundles. The translation-invariant Dirac operator is nothing but the Cauchy-Riemann operator $\partial/\partial x + i \partial/\partial y$.
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.)
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I think there is a slight mistake in the formulation of the question. $\mathbb{CP}^n$ is the homogeneous space $U(n+1)/(U(n) \times U(1))=SU(n+1)/G$ with $G= SU(n+1) \cap (U(n) \times U(1))$. The right formulation of question (2) is: is the spin structure on $\mathbb{CP}^n$ (for odd $n$, there is unique spin structure on $\mathbb{CP}^n$, see Charles Siegel's answer) $U(n+1)$-equivariant?
The answer is no, for a very elementary reason: if the spin structure were $U(n+1)$-equivariant, then it certainly were $U(n)$-equivariant, where $U(n)$ embeds into the product in the standard way. But the $U(n)$-action on $\mathbb{CP}^n$ has a fixed point and it is not too hard to see that the $U(n)$-representation on the tangent space to that fixed point is isomorphic to the standard representation of $U(n)$ on $\mathbb{C}^n$. So if the spin structure were equivariant, then the fixed-point representation has to be spin, which is of course wrong.
You can ask the same question for spheres (is the spin structure on $S^n$ $SO(n+1)$-equivariant), and the answer is again no. But the spin structure on $S^n$ is $Spin(n+1)$-equivariant; likewise the spin structure on $\mathbb{CP}^n$ will be equivariant under the double cover of $U(n)$.
What you can guess from these two examples is that the question has something to do with double covers (alias central extensions of your group by $\mathbb{Z}/2$). Here is the precise relation: $M$ a spin manifold, $s$ a spin structure (viewed as a double cover of the frame bundle of $M$), $G$ a topological group acting on M by diffeomorphisms. The spin structure defines a new group $G'$ and a surjective homomorphism $p:G' \to G$ with kernel. $G'$ consists of pairs $(f,t)$, $f \in G$ and $t$ is an isomorphism of spin structures $f^* s \to s$. The spin structure is equivariant under $G'$, and it is $G$-equivariant iff there is $q:G \to G'$, $pq=\operatorname{id}$. If $G$ is a simply-connected topological group, this is always the case, but otherwise not in general.
This discussion implies that the spin structure on $\mathbb{CP}^n$ is indeed $SU(n+1)$-equivariant, if it exists. Grassmannians and other homogeneous spaces can be dealt with in the same way.