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 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.
Best Answer
Let $L \to S^1\times S^1$ denote the line bundle you describe.
The restriction of $L$ to $S^1\times\{p\}$ is non-trivial, so $L$ is non-trivial and hence $w_1(L) \neq 0$ since real line bundles are classified up to isomorphism by their first Stiefel-Whitney class. Identifying $w_1(L) \in H^1(S^1\times S^1; \mathbb{Z}_2)$ requires some more work.
For each $p \in S^1$ we have inclusions $i_p : S^1 \to S^1\times S^1$ and $j_p : S^1 \to S^1\times S^1$ given by $i_p(q) = (q, p)$ and $j_p(q) = (p, q)$ respectively. We also have projections $\pi_i : S^1\times S^1 \to S^1$ for $i = 1, 2$ where $\pi_i(a_1, a_2) = a_i$. Note that $\pi_1\circ i_p = \operatorname{id}_{S^1}$ and $\pi_2\circ j_p = \operatorname{id}_{S^1}$. On the other hand, $\pi_2\circ i_p = c_p$ and $\pi_1\circ j_p = c_p$ where $c_p : S^1\to S^1$ is the constant map given by $c_p(q) = p$.
Let $\gamma \to S^1$ denote the non-trivial real line bundle on $S^1$. From your description of $L$, we have $i_p^*L \cong \gamma$ and $j_p^*L \cong \gamma$.
By the Künneth theorem, there is an isomorphism
\begin{align*} H^1(S^1; \mathbb{Z}_2)\oplus H^1(S^1; \mathbb{Z}_2) &\to H^1(S^1\times S^1; \mathbb{Z}_2)\\ (\alpha, \beta) &\mapsto \pi_1^*\alpha + \pi_2^*\beta. \end{align*}
In particular, $w_1(L) \in H^1(S^1\times S^1; \mathbb{Z}_2)$ is of the form $\pi_1^*\alpha + \pi_2^*\beta$ for some $\alpha$ and $\beta$. Now note that
\begin{align*} w_1(L) &= \pi_1^*\alpha + \pi_2^*\beta\\ i_p^*w_1(L) &= i_p^*\pi_1^*\alpha + i_p^*\pi_2^*\beta\\ w_1(i_p^*L) &= (\pi_1\circ i_p)^*\alpha + (\pi_2\circ i_p)^*\beta\\ w_1(\gamma) &= \operatorname{id}_{S^1}^*\alpha + c_p^*\beta\\ w_1(\gamma) &= \alpha. \end{align*}
Likewise
\begin{align*} w_1(L) &= \pi_1^*\alpha + \pi_2^*\beta\\ j_p^*w_1(L) &= j_p^*\pi_1^*\alpha + j_p^*\pi_2^*\beta\\ w_1(j_p^*L) &= (\pi_1\circ j_p)^*\alpha + (\pi_2\circ j_p)^*\beta\\ w_1(\gamma) &= c_p^*\alpha + \operatorname{id}_{S^1}^*\beta\\ w_1(\gamma) &= \beta. \end{align*}
Therefore
$$w_1(L) = \pi_1^*\alpha + \pi_2^*\beta = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma).$$
Now that we have identified $w_1(L)$ and real line bundles are classified up to isomorphism by their first Stiefel-Whitney class, we can find an expression for $L$. Since
$$w_1(L) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma) = w_1(\pi_1^*\gamma) + w_1(\pi_2^*\gamma) = w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma),$$
we see that $L \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$. For a proof of the last equality above, see the Lemma in this answer.