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:

A $3$-manifold, not necessarily closed, is parallelizable iff the first two Stiefel-Whitney classes $w_1, w_2$ vanish iff it admits an orientation and a spin structure.

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

A $4$-manifold, not necessarily closed, is parallelizable iff the characteristic classes $w_1, w_2, e, p_1$ all vanish iff it admits an orientation, a spin structure, and a string structure.

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.