Test for a $G$-torsor to be trivial? I just have a very short question, why is a $G$-torsor trivial precisely when it has a section?
 A: As RghtHndSd has pointed out, with a section you can define a global isomorphism.
Let's just recall, what I assume, is the definition of torsor you are using. Namely, let's let $X$ be a scheme, and $\mathscr{S}$ any of the usual sites on $X$ (e.g. $X_{\mathrm{Zar}},X_{\mathrm{\acute{e}t}},X_{\mathrm{fppf}}$,...). Also, let $G$ be an algebraic group over $X$. A $G$-torsor is then a sheaf of sets $\mathcal{F}$ on $\mathscr{S}$ together with an action $G\times\mathcal{F}\to\mathcal{F}$ such that

*

*For all $U$ an open in $\mathscr{S}$, the action $G(U)\times F(U)\to F(U)$ is simply transitive.

*For each $T$ an object of $\mathscr{S}$, there exists an open cover $\{U_i\to T\}$ such that $G(U_i)\ne\varnothing$.

Note that if $G(U)\ne\varnothing$, say $x\in G(U)$, then $g\mapsto g\cdot x$ is an isomorphism of sets $G(U)\to \mathcal{F}(U)$.
So, now, let's suppose that $F(X)\ne\varnothing$, again, let's say $x\in F(X)$. Then, we get an isomorphism $G(X)\to F(X)$ as before. But, note that $x\mid_T\in F(T)$ is a section in $F(T)$ for all $T\in\mathscr{S}$ (here $x\mid_T$ means the image under the map $F(X)\to F(T)$). One can then check that the map
$$G(T)\to \mathcal{F}(T):g\mapsto g\cdot x\mid_T$$
is actually a functorial isomorphism, so that $G=F$ as desired.

Here's a good example to keep in mind. Let $\mathscr{L}$ be a line bundle on $X$ (in any of the natural sites--they are all equivalent notions). Then, we naturally get a $\mathbf{G}_m$-torsor associated to $\mathscr{L}$. Let's call it $\mathcal{F}_{\mathscr{L}}$.
Then, one hope that $\mathcal{F}_{\mathscr{L}}$ is trivial (as a $\mathbf{G}_m$-torsor) if and only if $\mathscr{L}$ is trivial. But, as we have just seen, $\mathcal{F}_{\mathscr{L}}$ is trivial if and only if $\mathcal{F}_{\mathscr{L}}(X)\ne\varnothing$.
But, this is clear. Why? What is the torsor $\mathcal{F}_{\mathscr{L}}$? It's just the 'isom' sheaf
$$\mathrm{Isom}(\mathcal{O}_X,\mathscr{L}):U\mapsto \left\{\text{isomorphisms }\mathcal{O}_U\to\mathscr{L}\mid_U\right\}$$
Thus, we see trivially, definitionally, concretely why a line bundle $\mathscr{L}$ (or equivalently it's torsor $\mathcal{F}_{\mathscr{L}}$) is trivial if it has a global section--if $\mathrm{Isom}(\mathcal{O}_X,\mathscr{L})$ has a global section, then that literally means that there is an isomorphism
$$\mathcal{O}_X\to \mathscr{L}\mid_X=\mathscr{L}$$
I hope this helps!
