I'm trying to study line bundle over $S^2$. In this post was outlined the method based on clutching functions. But now I'm interesting in another approach.
For the sphere there is two maps : upper hemisphere and lower hemisphere with intersection as $[-\epsilon,\epsilon]\times S^1$. For the upper hemisphere and lower hemisphere its well-known that bundles over this spaces is trivial. (Any bundle over a contractible base is trivial). So to prove the fact that line bundle over $S^2$ is trivial we must create continuation of trivialization from upper hemisphere (for example) to the lower hemisphere through "border" $[-\epsilon,\epsilon]\times S^1$.
As I understand it is sufficient to continue trivialization from the "border" to the center of the "disk". (I think here it is possible to use a partition of unity, but I'm not sure).
I can't formalize this reasoning.
A trivialization of a real line bundle over an open set $U$ is essentially a (continuous) non-vanishing function $\sigma:U \to \mathbf{R}$. If $U$ is connected (e.g., an enlarged hemisphere), such a function has constant sign, say positive.
If the upper hemisphere is trivialized, you now have a positive function $\sigma$ on the boundary of the lower hemisphere. Extending the trivialization over the lower hemisphere amounts to extending $\sigma$ to a continuous, positive function in the lower hemisphere, which may be identified with the unit disk. As you note, one method is to construct a partition of unity.
This can be done explicitly: Choose radii $0 < r_1 < r_2 < 1$ and define functions in polar coordinates: $$ f_1(r, \theta) = \begin{cases} 0 & \text{if $r \leq r_1$}, \\ \frac{r - r_1}{r_2 - r_1} & \text{if $r_1 < r < r_2$}, \\ 1 & \text{if $r_2 \leq r$}, \end{cases} $$ and $f_2(r, \theta) = 1 - f_1(r, \theta)$. Viewing $\sigma$ as a function of $\theta$ alone, the function $\sigma(\theta) f_1(r, \theta) + f_2(r, \theta)$ is a continuous extension of $\sigma$ to a positive function in the disk.