How does a section of a bundle with fibre $\mathbb{P}^N$ give us a spin structure? In "Spin Structures on Manifolds" Milnor says that for a principal $SO(n)$ bundle $\pi:E\to X$ a spin structure can be defined as follows: Let $Spin(n)$ act on a sphere $S^N$ in such a way that the kernel $\mathbb{Z}_2$ of the cover $Spin(n) \to SO(n)$ acts freely, thus giving an action of the quotient $SO(n) $ on the quotient $\mathbb{P}^N$.
He considers the bundle with fibre $\mathbb{P}^N$ associated to the principal bundle $\pi:E\to X$ and says that a spin structure can be seen as a section of this bundle but he doesn't give any detail. How can we get a spin structure on $E$ from such a section? 
 A: Let $\lambda \colon Spin(n) \to SO(n)$ be the usual double-cover; let $\rho\colon Spin(n) \to SO(N+1)$ be a spinor representation, so that $ker(\lambda)\cong \mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim\ X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$  be a principal $SO(n)$-bundle over $X$, classified by $c\colon X\to BSO(n)$; finally let $\mathbb{P}_{SO}$ be the $\mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation. 
The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $\mathbb{P}_{SO}$?"
There is a model of $BSpin(n)$ as a $\mathbb{P}^\infty \sim B(\mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $\mathbb{P}^\infty$ in a way induced by $\rho$, and the inclusion $\mathbb{P}^N\to\mathbb{P}^\infty$ induces an inclusion $\mathbb{P}_{SO} \to c^*BSpin(n)$. From a section $\sigma\colon X\to \mathbb{P}_{SO}$ we can form the composition 
$$ \tilde{c}\colon X \to \mathbb{P}_{SO} \to c^*BSpin(n) \to BSpin(n) $$
which is a lift of $c$ along $B\lambda$. Then $\tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.
The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $\mathbb{P}_{SO}$?"
I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $B\lambda\colon BSpin(n) \to BSO(n)$, and then from lifts of classifying maps to sections of $\mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.
Step 1: Spin structures correspond to lifts of classifying maps.
The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $B\lambda\colon BSpin(n) \to BSO(n)$ (this is a reasonably common interpretation). Specifically let $c\colon X\to BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) \to X$ and consider the diagram 
$\require{AMScd}$
\begin{CD}
    P_{SO} @>>> ESO(n) @< E\lambda << ESpin(n) \\
    @V  V V @VV \pi_{SO} V  @VV \pi_{Spin} V\\
    X @>c>> BSO(n) @< B\lambda << BSpin(n)
\end{CD}
The functorial map $E\lambda\colon ESpin(n) \to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor. 
Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $B\lambda$. 
Proof Sketch. If $\tilde{c}\colon X\to BSpin(n)$ is a lift of $c$ along $B\lambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = \tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $P\to P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $\tilde{c}\colon X\to BSpin(n)$ and we need to show that $c = B\lambda\circ\tilde{c}$, at least up to homotopy. The claim is that $\tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $\tilde{c}^*ESpin(n)\cong P$ and then $B\lambda \circ \tilde{c} \sim c$ by classifying space theory.
Step 2: Lifts of classifying maps correspond to sections of the projective space bundle
Let $\mathbb{P}_{SO}$ be the bundle with fibre $\mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim\ X$.
Proposition 2. Homotopy classes of lifts of $c$ along $B\lambda$ correspond to homotopy classes of sections of $\mathbb{P}_{SO}$.
Proof sketch.  This is essentially due to the following fact: If $K\to G \to H$ is a short exact sequence of groups then $BK \to BG \to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG \to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(\mathbb{Z}/2)\sim \mathbb{P}^\infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $\mathbb{P}^\infty$ in the same way. The inclusion $\mathbb{P}^N \to \mathbb{P}^\infty$ induces a fibre-wise inclusion $\mathbb{P}_{SO} \to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $\mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $B\lambda$, finishing the proof sketch.
(Supplemental: The Borel Construction)
Definition.  Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $X\times Y$ is defined by $(x, y)\cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as
$$X\times_{G} Y := (X \times Y)/G.$$
(Equivalently you could take the quotient by the relation $(xg,y) \sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $G\backslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $\varphi\colon G\to Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
$$E\cong P\times_G F.$$ 
I'm going to use the following lemma to switch from a $(SO(n),\mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:
Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $X\times_G Y$ is homeomorphic to $X \times_{G/K}(Y/K)$.
(The idea is that whatever acts trivially on the left side can be eaten up on the right side.) 
Now suppose $K\to G \to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $G\to H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EH\times E$ is free. Then
$$ BG = EH \times_{G} E \cong EH\times_{G/K} (E/K) \cong EH \times_H BK$$
That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.
In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^\infty$ where $\mathbb{Z}/2$ acts freely, so then
$$ BSpin(n) = ESO(n)\times_{Spin(n)} S^\infty \cong ESO(n) \times_{SO(n)} \mathbb{P}^\infty $$ 
Moreover the map $ESO(n)\times_{SO(n)} \mathbb{P}^N \to BSpin(n)$ is as connected as the map $\mathbb{P}^N \to \mathbb{P}^\infty$
