$\newcommand{\O}{\mathcal{O}}$
$\newcommand{\F}{\mathcal{F}}$
The way you get a locally free sheaf of rank $n$ from a $GL(n)$-torsor $P$ is by twisting the trivial rank $n$ bundle $\O^n$ (which has a natural $GL(n)$-action) by the torsor. Explicitly, the locally free sheaf is $\F=\O^n\times^{GL(n)}P$, whose (scheme-theoretic) points are $(v,p)$, where $v$ is a point of the trivial bundle and $p$ is a point of $P$, subject to the relation $(v\cdot g,p)\sim (v,g\cdot p)$. Conversely, given a locally free sheaf $\F$ of rank $n$, the sheaf $Isom(\O^n,\F)$ is a $GL(n)$-torsor, and this procedure is inverse to the $P\mapsto \O^n\times^{GL(n)}P$ procedure above. (Note: I'm identifying spaces over the base $X$ with their sheaves of sections, both for regarding $Isom(\O^n,\F)$ as a torsor and for regarding $\O^n\times^{GL_n}P$ as a locally free sheaf.)
Similarly, if you have a group $G$ and a representation $V$, then you can associate to any $G$-torsor $P$ a locally free sheaf of rank $\dim(V)$, namely $V\times^G P$. But I don't know of a characterization of which locally free sheaves of rank $\dim(V)$ arise in this way.
Operations with the locally free sheaf (like taking top exterior power, or any other operation which is basically defined fiberwise and shown to glue) correspond to doing that operation with the representation $V$, so I think you're right that in the case of $SL(n)$ you get exactly those locally free sheaves whose top exterior power is trivial (since $SL(n)$ has no non-trivial $1$-dimensional representations).
While the Zariski topology has its limitations, it amazes me how well it does work. A few brief points in its defense:
It's easy to define. In the classical case of affine spaces over a field, it's the weakest topology for which points are closed and polynomials are continuous.
It can be used to give precise meaning to the word "generic" or "general",
as in "a general matrix is diagonalizable, therefore to prove Cayley-Hamilton it suffices..."
For coherent sheaves, it's the right topology; cohomology works as expected.
At a more sophisticated level, cohomology is upper semicontinuous in the
Zariski topology, and this is very important for many arguments.
So for the younger generation out there who are thinking of doing away with it: please
don't!
Best Answer
One way to think about $H^1(A)$ is to use the long exact sequence not as a property of cohomology, but outright as a definition. That is, given an exact sequence of sheaves, $$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ then $H^1(A)$ is measuring the obstruction of global sections to be exact: $$ 0\rightarrow \Gamma(A)\rightarrow \Gamma(B)\rightarrow \Gamma(C)\rightarrow H^1(A)$$ In words, $H^1$ is `measuring the failure of $\Gamma$ to preserve surjectivity'. If you want this idea to actually define $H^1(A)$, you have to be careful to chose $B$ so that $H^1(B)=0$. But, as far as intuition goes, this works pretty well for me.
A demonstrative example of this, at least for me, is to think about the the complex variety $\mathbb{P}^1$, with $\mathcal{O}$ the structure sheaf and $\mathbb{C}_p$ the skyscraper sheaf at a point. Then there is a surjective map (it is surjective because it is surjective on stalks): $$ \mathcal{O}\rightarrow\mathbb{C}_p$$ which has kernel $\mathcal{O}(-1)$, the twisted structure sheaf. This whole short exact sequence can be twisted by $(-1)$, noting that twisting a skyscraper sheaf $\mathbb{C}_p$ gives an isomorphic sheaf $\mathbb{C}_p(-1)$ (which I identify with the original sheaf): $$0\rightarrow \mathcal{O}(-2)\rightarrow \mathcal{O}(-1)\rightarrow \mathbb{C}_p\rightarrow 0$$
On global sections, we then get $$ 0\rightarrow \Gamma(\mathcal{O}(-2))\rightarrow \Gamma(\mathcal{O}(-1))\rightarrow \Gamma(\mathbb{C}_p)\rightarrow H^1(\mathcal{O}(-2))$$ We know that $\Gamma(\mathcal{O}(-1))=0$ and $\Gamma(\mathbb{C}_p)=\mathbb{C}$, so the middle arrow is no longer surjective. Hence, $H^1(\mathcal{O}(-2))$ must contain at least $\mathbb{C}$ (in fact, it is exactly $\mathbb{C}$, since $H^1(\mathcal{O}(-1))=0$).
Higher cohomology may be also thought of this way: $H^{i+1}$ measures the failure of $H^i$ to preserve surjective maps. However, I don't find this very useful for thinking about higher cohomology, since it would need that I somehow understand lower cohomology much better.