The answer is more straightforward than what I was saying in the comments above, though the general principle of what I was saying holds. Let $n$ denote the dimension of $M$.
Recall that the Stiefel-Whitney classes are characterized by a set of four axioms. The normalization axiom says that $w_1$ of the unorientable line bundle over $S^1$ is the generator of $H^1(S^1, \mathbb{Z}/2)$ and that $w_1$ of the orientable line bundle is $0 \in H^1(S^1, \mathbb{Z}/2)$.
Fix now a loop $\gamma \colon S^1 \to M$. The question of whether $TM$ is orientable over $S^1$ corresponds to whether the determinant bundle $\Lambda^n TM$ is trivial or not when restricted to $\gamma$. Naturality of the Stiefel-Whitney class gives
\[
< w_1( \gamma^* \Lambda^n TM), [S^1] > = < \gamma^* w_1(\Lambda^n TM), [S^1]>
\]
where this pairing is between cohomology/homology of $S^1$.
Now, the LHS is equal to $1$ iff $\gamma^*(\Lambda^n TM)$ is the non-trivial bundle, i.e. iff this loop is non-orientable. The RHS is equal to $ < w_1(\Lambda^n TM), \gamma_*[S^1]>$. This shows that $w_1$ exhibits the property you claim.
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.
Best Answer
Initially, Stiefel-Whitney classes are only defined for smooth manifolds (if the manifold is not smooth, it doesn't have a tangent bundle). However, Wu's theorem states that $w = \operatorname{Sq}(\nu)$ and we can use this as a definition of Stiefel-Whitney classes for topological manifolds; in particular, $w_4(M)$ is defined even if $M$ does not admit a smooth structure or even a PL structure.
Freedman's Theorem states every non-degenerate bilinear form $b$ arises as the intersection of a closed, simply connected four-manifold. Moreover, if $b$ is even, there is a unique such manifold up to homeomorphism, and if $b$ is odd, there are two up to homeomorphism and they are distinguished by the Kirby-Siebenmann invariant.
As $E_8$ is an even form, there is a unique (up to homeomorphism) closed, simply connected four-manifold with intersection form $E_8$; we denote it by $M_{E_8}$.
As $(1)$ is an odd form, there are (up to homeomorphism) two closed, simply connected four-manifolds with intersection form $(1)$. One is $\mathbb{CP}^2$ which has zero Kirby-Siebenmann invariant (it admits a smooth structure and hence a PL structure). We denote the other manifold by $*\mathbb{CP}^2$ and note that it has non-trivial Kirby-Siebenmann invariant.
Recall that for a closed smooth four-manifold $M$, the Stiefel-Whitney number $\langle w_4(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic $\chi(M)$; this is Corollary $11.12$ of Milnor and Stasheff's Characteristic Classes. The same is true for topological manifolds (using the definition of Stiefel-Whitney classes outlined above), see this question.
With all of this in mind, we can now see that $\operatorname{ks}(M)$ and $w_4(M)$ are unrelated as the following table demonstrates.
$$ \begin{array}{c|cc} M & \operatorname{ks}(M) & w_4(M)\\ \hline S^4 & 0 & 0 \\ \mathbb{CP}^2 & 0 & \neq 0 \\ M_{E_8} & \neq 0 & 0 \\ *\mathbb{CP}^2 & \neq 0 & \neq 0 \end{array} $$
If the four-manifold $M$ admits a PL structure, then $\operatorname{ks}(M) = 0$, but the converse is not true. For example, $M_{E_8}\# M_{E_8}$ has trivial Kirby-Siebenmann invariant but it does not admit a PL structure (every PL manifold of dimension less than or equal to $7$ is smoothable, but $M$ is not smoothable by Donaldson's theorem). The class $w_4(M)$ has nothing to do with the obstructions to admitting a PL structure (both $S^4$ and $\mathbb{CP}^2$ admit PL structures).