Sorry to resurrect, but we leave a $($detailed$)$ proof here that $TM$ has the structure of an oriented $2n$-manifold, even if the $n$-manifold $M$ is non-orientable.
Let $\{(U_\alpha, \phi_\alpha)\}_{\alpha \in A}$ be a smooth atlas of $M$, and let $V_\alpha = \phi_\alpha(U_\alpha) \subset \mathbb{R}^n$. Then $(\phi_\alpha)_*: TU_\alpha \to TV_\alpha = V_\alpha \times \mathbb{R}^n$ is a homomorphism $($having inverse $(\phi_\alpha^{-1})_*$$)$. Moreover, the sets $TU_\alpha$ cover $TM$ and the transition maps$$t_{\alpha\beta} = (\phi_\alpha)_* \circ \left(\phi_\beta^{-1}\right)_* = \left(\phi_\alpha \circ \phi_\beta^{-1}\right)_*: V_\beta \times \mathbb{R}^n \to V_\alpha \times \mathbb{R}^n$$are orientation preserving. Let $x_1, \dots, x_n$ be coordinates on $V_\beta$, $x_{n+1}, \dots, x_{2n}$ be coordinates on the left copy of $\mathbb{R}^n$, $y_1, \dots, y_n$ be coordinates on $V_\alpha$, and $y_{n+1}, \dots, y_{2n}$ be coordinates on the right copy of $\mathbb{R}^n$. Note that $(y_1, \dots, y_n) = \phi_\alpha\left(\phi_\beta^{-1}(x_1, \dots, x_n)\right)$ does not depend on $x_{n+1}, \dots, x_{2n}$, so the Jacobian matrix $\left({{\partial y_i}\over{\partial x_j}}\right)$ of $t_{\alpha\beta}$ has all zeros in the upper right quadrant. It follows that$$\det\left({{\partial y_i}\over{\partial x_j}}\right)_{i,j = 1}^{2n} = \det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = 1}^n \det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = n +1}^{2n}.$$The first of these submatrices is the usual Jacobian of $\phi_\alpha \circ \phi_\beta^{-1}$. The second is the Jacobian of the linear transformation $\left(\phi_\alpha \circ \phi_\beta^{-1}\right)_{*,\, (x_1, \dots, x_n)}$. Therefore$$\det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = 1}^{2n} = \det\left(\left(\phi_\alpha \circ \phi_\beta^{-1}\right)_{*,\, (x_1, \dots, x_n)}\right)^2 > 0.$$This proves $\{(TU_\alpha, (\phi_\alpha)_*)\}_{\alpha \in A}$ is an oriented atlas for $TM$.
There are some confusions here I should clear up before answering.
1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;\Bbb Z/2)$-many of them, and a spin manifold requires one such choice.
2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) \in H^i(M;\Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < \dim M$, then so does $M \# M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{\dim M}(M; \Bbb Z/2) \cong \Bbb Z/2$; they are labeled by partitions of $\dim M$.
3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.
As for your actual question, the point is that we have a natural isomorphism $T(\partial W) \oplus \Bbb R \cong TW\big|_{\partial W}$. The easy claim is that if $W$ is spinnable, then $\partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(\partial W)$, where $j: \partial W \to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $\partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.
To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E \oplus \Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E \oplus \Bbb R$ (via the natural map $\text{Spin}(n) \to \text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;\Bbb Z/2)$.
This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $\partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;\Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.
Best Answer
I don't know of a totally elementary proof of this result, but here is some context for it. More generally, an $n$-manifold $M$ (without boundary) has a classifying map $f : M \to BO(n)$ for its tangent bundle. Knowing when $M$ is parallelizable is equivalent to knowing when $f$ is null-homotopic. There is a general machine for doing this involving lifting $f$ higher and higher through the stages of the Whitehead tower of $BO(n)$, and it tells us that the complete set of obstructions to solving this problem are a set of cohomology classes in $H^k(M, \pi_k(BO(n)), k \le n$, each of which is well-defined provided that the previous one vanishes, such that $f$ is null-homotopic iff all of the classes vanish.
The construction of the first such class goes like this. Assume for simplicity that $M$ is connected. The first question is whether $f$ lifts to the universal cover of $BO(n)$, which is true iff $f$ induces the zero map on $\pi_1$ by standard covering space theory. Now, $\pi_1(BO(n)) \cong \mathbb{Z}_2$, and the induced map on $\pi_1$ gives a homomorphism $\pi_1(M) \to \mathbb{Z}_2$ which corresponds precisely to the first Stiefel-Whitney class $w_1$. This class vanishes iff $f$ lifts to the universal cover of $BO(n)$, which is $BSO(n)$, iff $M$ is orientable.
Now we want to try lifting $f$ to the $2$-connected cover of $BSO(n)$; this is analogous to the universal cover but involves killing $\pi_2(BSO(n)) \cong \mathbb{Z}_2$ (for $n \ge 3$) instead of $\pi_1$. Whether this is possible is controlled by the map
$$BSO(n) \to B^2 \mathbb{Z}_2$$
inducing an isomorphism on $\pi_2$. This is equivalently a universal characteristic class in $H^2(BSO(n), \mathbb{Z}_2)$ which turns out to be precisely the second Stiefel-Whitney class $w_2$. This class vanishes iff $f$ lifts to the $2$-connected cover of $BSO(n)$, which is $BSpin(n)$, iff $M$ has a spin structure.
The first surprise in this story is that (when $n \ge 3$) $BSpin(n)$ also turns out to be the $3$-connected cover; in other words, $\pi_3(BSpin(n)) = 0$, so the next step of this story involves $\pi_4$ and can be ignored for $3$-manifolds. If $M$ is a $3$-manifold, not necessarily compact, admitting both an orientation and a spin structure, then the classifying map of its tangent bundle lifts to a map $M \to BSpin(3)$, but since the latter is $3$-connected any such map is nullhomotopic. So:
To give an indication of the generality of this machinery, for $4$-manifolds the next step involves computing $\pi_4(BSpin(4)) \cong \mathbb{Z}^2$ and then lifting to the $4$-connected cover of $BSpin(4)$, which is called $BString(4)$. This says that the next obstruction to a $4$-manifold with an orientation and a spin structure being parallelizable is a pair of cohomology classes in $H^4(M, \mathbb{Z})$ which I believe turn out to be the Euler class $e$ and the first fractional Pontryagin class $\frac{p_1}{2}$ (a certain characteristic class of spin manifolds that when doubled gives the Pontryagin class $p_1$) respectively. Since $H^4(M, \mathbb{Z})$ is always torsion-free for a $4$-manifold, the conclusion is that
And for $5$-manifolds and higher we really need to talk about $\frac{p_1}{2}$ and not just $p_1$.
The second surprise in this story is that for closed $3$-manifolds the condition that $w_2$ vanishes is redundant: it is implied by orientability by a standard computation with Wu classes. In other words, closed orientable $3$-manifolds automatically admit spin structures. I don't have a good intuitive explanation of this; it comes from a relationship between the Stiefel-Whitney classes, Steenrod operations, and Poincaré duality that I don't understand very well.