Spin structures and the second Stiefel-Whitney class are themselves not particularly simple, so I don't know what kind of an answer you're expecting. Here is an answer which at least has the benefit of being fairly conceptual.
First some preliminaries. Recall that a real vector bundle of rank $n$ on a space is the same thing as a principal $\text{GL}_n(\mathbb{R})$-bundle (namely its frame bundle) and that principal $G$-bundles are classified by maps into the classifying space $BG$. In particular, a smooth manifold $M$ of dimension $n$ has a tangent bundle which has a classifying map $M \to B\text{GL}_n(\mathbb{R})$. Additional information allows us to reduce the structure group of this classifying map as follows:
- If $M$ is equipped with a Riemannian metric then the transition maps for the tangent bundle can be chosen to lie in $\text{O}(n) \subseteq \text{GL}_n(\mathbb{R})$, so we get a classifying map $M \to B \text{O}(n)$.
- If $M$ is in addition equipped with an orientation then (possibly by definition, depending on your definition of an orientation) the transition maps for the tangent bundle can in addition be chosen to lie in $\text{SO}(n) \subseteq \text{O}(n)$, so we get a classifying map $M \to B \text{SO}(n)$.
- If $M$ is in addition equipped with a spin structure then (probably by definition, depending on your definition of a spin structure) the above classifying map can be lifted to a classifying map $M \to B \text{Spin}(n)$.
Now, what does this have to do with the second Stiefel-Whitney class? First let me tell a simpler story about the first Stiefel-Whitney class. The first Stiefel-Whitney class is a cohomology class $w_1 \in H^1(B\text{O}(n), \mathbb{F}_2)$ giving a characteristic class for $\text{O}(n)$-bundles which vanishes iff those bundles can be reduced to $\text{SO}(n)$-bundles. Why?
One reason is the following. $w_1$ can be regarded as a homotopy class of maps $B \text{O}(n) \to B \mathbb{Z}_2$ (where I use $\mathbb{Z}_2$ to mean the cyclic group of order $2$). Now, it's known that any such map comes from a homotopy class of maps $\text{O}(n) \to \mathbb{Z}_2$, and there's an obvious candidate for such a map, namely the determinant. This gives an exact sequence
$$1 \to \text{SO}(n) \to \text{O}(n) \to \mathbb{Z}_2 \to 1$$
which, after applying the classifying space functor, gives a homotopy fibration
$$B \text{SO}(n) \to B \text{O}(n) \xrightarrow{w_1} B \mathbb{Z}_2$$
exhibiting $B \text{SO}(n)$ as the homotopy fiber of the first Stiefel-Whitney class.
The homotopy fiber of a map between (pointed) spaces is analogous to the kernel of a map between groups; in particular, if $w : B \to C$ is a map of groups, then a map $f : A \to B$ satisfies $w \circ f = 0$ if and only if $f$ factors through a map $A \to \text{ker}(w)$. The same kind of thing is happening here: a classifying map $f : M \to B \text{O}(n)$ satisfies that $w_1 \circ f$ is homotopic to a constant map if and only if it factors up to homotopy through the homotopy fiber $M \to B \text{SO}(n)$.
Now the reason I gave such a sophisticated description of orientations is that the story for spin structures is completely parallel. Namely, the second Stiefel-Whitney class is a cohomology class $w_2 \in H^2(B\text{SO}(n), \mathbb{F}_2)$ which can be regarded as a homotopy class of maps $B \text{SO}(n) \to B^2 \mathbb{Z}_2$. You can produce such classes by applying the classifying space functor to a homotopy class of maps $\text{SO}(n) \to B \mathbb{Z}_2$, or equivalently a cohomology class in $H^1(\text{SO}(n), \mathbb{F}_2)$, and there's a natural candidate for such a class, namely the cohomology class classifying the nontrivial double cover $\text{Spin}(n) \to \text{SO}(n)$. This also turns out to imply that we get a homotopy fibration
$$B \text{Spin}(n) \to B \text{SO}(n) \xrightarrow{w_2} B^2 \mathbb{Z}_2$$
exhibiting $B \text{Spin}(n)$ as the homotopy fiber of the second Stiefel-Whitney class, and if you believe this then it again follows from the universal property of the homotopy fiber that a map $f : M \to B \text{SO}(n)$ lifts to a map $M \to B \text{Spin}(n)$ iff $w_2 \circ f$ is homotopic to a constant map.
(This homotopy fibration is a "delooping" of the more obvious homotopy fibration $B \mathbb{Z}_2 \to B \text{Spin}(n) \to B \text{SO}(n)$ coming from the short exact sequence $1 \to \mathbb{Z}_2 \to \text{Spin}(n) \to \text{SO}(n) \to 1$.)
This argument can be continued all the way up the Whitehead tower of $B \text{O}(n)$; the next step is a string structure, etc.
There are some confusions here I should clear up before answering.
1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;\Bbb Z/2)$-many of them, and a spin manifold requires one such choice.
2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) \in H^i(M;\Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < \dim M$, then so does $M \# M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{\dim M}(M; \Bbb Z/2) \cong \Bbb Z/2$; they are labeled by partitions of $\dim M$.
3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.
As for your actual question, the point is that we have a natural isomorphism $T(\partial W) \oplus \Bbb R \cong TW\big|_{\partial W}$. The easy claim is that if $W$ is spinnable, then $\partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(\partial W)$, where $j: \partial W \to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $\partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.
To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E \oplus \Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E \oplus \Bbb R$ (via the natural map $\text{Spin}(n) \to \text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;\Bbb Z/2)$.
This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $\partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;\Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.
Best Answer
A good reference for a lot of this is "Characteristic Classes" by Milnor and Stasheff. I can definitely answer 2, 3, 4 and partially answer 1.
1) As Peter points out in your comments, every closed orientable $3$-manifold is paralellizable. (You can use Wu's Formula (Theorem 11.4 in Milnor-Stasheff) to show all the Stiefel-Whitney numbers vanish, and then by work of Thom this is equivalent to being orientedly null-bordant since there are no Pontryagin classes.) In fact any counterexample would have to be non-compact but I don't know one.
2) No, the signature is an oriented bordism invariant, meaning that if $M^{4k}$ and $N^{4k}$ are orientedly bordant then $\sigma(M) = \sigma(N)$, in particular if $M\sim_{SO} S^4$ then $\sigma(M)=0$. The signature is one of the most computable ways of showing a manifold is not null-bordant, and in fact dimension $4$ is particularly nice because $\sigma\colon \Omega^{SO}_4 \to \mathbb{Z}$ is an isomorphism. Be careful however because there are many manifolds with vanishing signature which are not null-bordant, such as $(\mathbb{C}P^2 \times \mathbb{C}P^2) \# \overline{\mathbb{C}P^4}$.
3) Yes, it's known that the spin bordism group $\Omega^{spin}_1$ is $\mathbb{Z}/2$, where the two classes are represented by spin structures on $S^1$. $S^1$ is orientedly bordant to itself via $S^1\times I$ but there is no spin bordism between $(S^1,P)$ and $(S^1, P')$ where $P$ and $P'$ are the two different spin structures. I don't know if there is a proof of this result which is both clean and elementary, but I wrote something up as an asnwer to this question.
4) No, every orientable surface is null-bordant, by the classification of surfaces, i.e. $w_1(TM) = 0$ means that $M$ is null-bordant.