An alternative description of the first Stiefel-Whitney class

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.

If you have a triangulated manifold then it is orientable when its n simplices can be given signs (+/-1) so that the signed sum is an integer cycle.

If the manifold is not orientable then any signed sum of the n simplices is not a cycle and it turns out that what goes wrong is that some of the n-1 simplices in the boundary get counted twice rather than cancelling each other out.

This boundary which is a sum of doubles of n-1 simplices can be divided by two to give a 2 torsion Z homology class. This n-1 dimensional cycle represents the obstruction to orienting the manifold. I would strongly suspect that the first Stiefel Whitney class of the tangent bundle is the Poincare dual to this homology class. Check it out.