An alternative description of the first Stiefel-Whitney class I've heard this description of the first Stiefel-Whitney class of a vector bundle but I don't know why this is true. Can anyone help me with this, please?
The first Stiefel-Whitney class of a vector bundle is an element in the first cohomology group of the base space $H^1(B,Z_2)$. This element can be seen as a map $w_1 : \pi_1(B) \rightarrow Z_2$
(because of the universal coefficient theorem and the fact that $H_1$ is the abelianization of $\pi_1$), so it assigns to each loop based at a fixed point an element of $Z_2$. Now this number is $0$ if and only if that loop is orientation preserving (locally we have an orientation and when we go around a loop and come back to our first place, we can ask if orientation has changed or not)

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.
