Given a foliated manifold $M$, we have a splitting $TM \cong T\mathcal F \oplus N \mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T \mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $\eta$ is a real line bundle on $Y$, $E = \eta \oplus \eta \oplus \Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(\eta)^2$. So if the cup-square map $H^1(Y;\Bbb Z/2) \to H^2(Y;\Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $\eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;\Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;\Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.