Now that you have all the Stiefel-Whitney classes written down, the hard part is over. To compute Stiefel-Whitney numbers, recall that these are, by definition, obtained in the following way.
Start with a partition of $4$, that is, a sum of a bunch of positive numbers which give $4$. Here are all five of the options: $1+1+1+1,\, 1+1+2,\, 1+3,\, 2+2,\,$ and $4$.
For each choice, form the corresponding product of Stiefel-Whitney class \begin{align*} 1+1+1+1 &\leftrightarrow w_1 \cup w_1 \cup w_1 \cup w_1 \\ 1+1+2 &\leftrightarrow w_1\cup w_1\cup w_2\\ 1+3 &\leftrightarrow w_1\cup w_3\\ 2+2 &\leftrightarrow w_2\cup w_2 \\ 4&\leftrightarrow w_4\end{align*}
The point of a partition is that all the cup products on the right land in $H^4(P^2\times P^2;\mathbb{Z}/2)$. Since every manifold has an orientation class mod $2$, we can pair the element on the right with the orientation class and get a number mod $2$ out. These numbers mod $2$ are the Stiefel-Whitney numbers.
By Poincare duality, the orientation class is the dual of the unique element in $H^4(P^2\times P^2,\mathbb{Z}/2)$, that is, it's the dual of $a^2 b^2$. Hence, computing all the Stiefel-Whitney numbers is the same as computing all the above cup products (using the relations $a^3 = b^3 = 0$), and then counting, mod $2$, the number of occurrences of $a^2 b^2$.
Doing this (while supressing the cup product sign) gives \begin{align*} (w_1)^4 &= (a+b)^4 & &= 0\\ (w_1)^2 w_2 &= (a+b)^2(a^2 + b^2 + ab) & &= 0 \\ w_1 w_3 &= (a+b)(ab^2 + a^2 b) & &= 0\\ (w_2)^2 &= (a^2+b^2+ab)^2 & &= a^2 b^2\\ w_4 &= a^2b^2 & &= a^2 b^2.\end{align*}
(Note that the computations are considerable eased by noting we're working mod $2$ so $(a+b)^2 = a^2 + b^2$.)
From this calculation, we see that three of the Stiefel-Whitney numbers are $0$ (mod $2$) while the other two are $1$ (mod $2$).
The following is from Lecture notes of Professor Farrell. A hopefully working link is here(page 95):
https://www.dropbox.com/s/80n4wd6xctpe6yr/Characteristic%20Classes%20%28Sparkie%20%E7%9A%84%E5%86%B2%E7%AA%81%E5%89%AF%E6%9C%AC%202014-02-19%29.pdf
We now reach proposition 7. Let $E$ be a complex vector bundle, then the mod 2 reduction of the total Chern class of $E$ is the total Stiefel-Whitney class of $E$. Since there is no Chern classes in odd dimensions, we have there is no Stiefel-Whitney classes in odd dimension as well.
$\textbf{Proof}$
Suppose $E$ is a line bundle. Then we know $w_{0}(L)=1, w_{1}(L)=0,w_{2}(L)=e_{\mathbb{Z}_{2}}(L)=\phi(e(L))$. On the other hand we know $c_{0}(L)=1, c_{1}(L)=e(L)$. So this verified it for line bundles.
For sum of line bundles we have
$$
E=\oplus^{n}_{i=1}L_{i}
$$
So the total Chern class is
$$
c(E)=c(L_{1}) \cup \cdots \cup c(L_{n})\mapsto_{\phi}\omega(E)=\omega(L_{1}) \cup \cdots \cup \omega(L_{n})
$$
We are now going to use splitting principle. But there is a subtle point. We now discuss it.
By the splitting principle there exist
$$
f:\mathcal{B}\rightarrow B
$$
such that
$$
f^{*}(E)=\oplus_{i=1}^{n}L_{i}
$$ and
$$
f^{*}:H^{*}(B,R)\rightarrow H^{*}(\mathcal{B},R)
$$
is monic.
Therefore by naturality and the fact $f^{*}$ is monic with respect to $\mathbb{Z}_{2}$ coefficients.
$$
\phi(f^{*}(c(E)))=f^{*}(\phi(c(E)))=f^{*}(\omega(E))
$$
Best Answer
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.