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.
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
Assume that $M$ is oriented throughout. Recall that $M$ has a $\text{Spin}^c$ structure iff the third integral Stiefel-Whitney class $\beta w_2 = W_3 \in H^3(M, \mathbb{Z})$ is trivial. Actually more is true: $\text{Spin}^c$ structures on $M$ are in bijection with trivializations of $W_3$, which are a torsor over $H^2(M, \mathbb{Z})$. So we get a functorial way to associate a $\text{Spin}^c$ structure to an almost complex structure for every choice of nullhomotopy of the composite map
$$BU(n) \to BSO(2n) \xrightarrow{W_3} B^3 \mathbb{Z}.$$
Isomorphism classes of nullhomotopies are a torsor over $H^2(BU(n), \mathbb{Z}) \cong \mathbb{Z}$. I don't know in what sense there's a canonical one, but once you pick one you don't have to make any further choices, and in particular you don't have to make any further choices involving $M$. (Edit #2: But see below.)
As for almost complex structures, there is a canonical fiber bundle
$$SO(2n)/U(n) \to BU(n) \to BSO(2n)$$
and almost complex structures on $M$ correspond to sections of the pullback of this bundle along the classifying map $M \to BSO(2n)$ of the tangent bundle. This admits a map of bundles to the corresponding bundle
$$B^2 \mathbb{Z} \to B \text{Spin}^c(2n) \to BSO(2n)$$
describing $\text{Spin}^c$ structures. The map $SO(2n)/U(n) \to B^2 \mathbb{Z}$ describes a canonical line bundle on the space of linear complex structures on $\mathbb{R}^{2n}$, namely the complex determinant bundle.
This tells us that we already can't expect all $\text{Spin}^c$ structures to come from almost complex structures when $n = 1$: here $SO(2)/U(1)$ is contractible, so there is in a very strong sense a unique almost complex structure on an oriented surface, but $\text{Spin}^c$ structures are a torsor over $H^2(-, \mathbb{Z})$, which can be nontrivial, e.g. on an oriented closed surface.
Edit #2: I believe that there is in fact a distinguished nullhomotopy of the composite map $BU(n) \xrightarrow{W_3} B^3 \mathbb{Z}$, as follows.
We now need the additional assumption that $M$ is Riemannian (although the choice of Riemannian metric won't matter). Then a conceptual description of $W_3 \in H^3(M, \mathbb{Z})$ is that it classifies the bundle of complex Clifford algebras $\text{Cliff}(T_x(M)) \otimes \mathbb{C}$ up to Morita equivalence. Trivializations of this bundle up to Morita equivalence correspond to Clifford module bundles with fiber the unique irreducible representation of $\text{Cliff}(2n) \otimes \mathbb{C}$ (complex spinor bundles).
If $M$ has an almost complex structure, then a distinguished complex spinor bundle can be constructed out of the complex exterior algebra of the tangent bundle. See Exercise 2.1.37 in Freed's Geometry of Dirac Operators for details. The torsor structure over $H^2(M, \mathbb{Z})$ comes from tensoring this bundle with complex line bundles on $M$, and the action of $H^2(BU(n), \mathbb{Z})$ comes from tensoring this bundle with powers of the canonical bundle. It's possible one might want to do this with a particular power; I'm not sure what's going on here exactly, but see Exercise 2.1.54 in Freed.
Edit #3: Now that the third question has been clarified, I don't know the answer off the top of my head. In the case of closed oriented surfaces we know that there's a $\mathbb{Z}$'s worth of $\text{Spin}^c$ structures. The unique almost complex structure gives rise to one, and the $2^{2g}$ spin structures also give rise to one (not depending on the spin structure, basically because the Bockstein homomorphism $H^1(M, \mathbb{Z}_2) \to H^2(M, \mathbb{Z})$ is zero in this case). But I don't know if they're the same one; they might differ by a power of the canonical bundle or something (or not, depending on your convention for trivializing $W_3$ as above). In summary,