Let $X$ be a smooth variety (or scheme) over an algebraically closed field $k$. Then we always regard it as a manifold. I'm wondering if this analog is literally? For example:
Is there an open covering $U_i$ of $X$ such that every $U_i$ is isomorphic to an affine open of $\mathbb A^n$?
Could we use local coordinates, just like that in differential geometry, to do concrete computation?
Is there any theorem to guarantee everything is okay?
My question is from the following proof. It seems that the author totally regards $S$ as a manifold and choose local coordinates to compute directly.
I would like to point out that the book you're taking the image from, Beauville's book on comples surfaces (of which I recognize the font and content), only deals with smooth projective varieties over the complex numbers. These varieties are always compact complex manifolds and so there's no problem with what Beauville is doing here.
As for your question, I belive the answer is yes (as far as smooth projective varieties are concerned). You can always find an affine cover (by intersecting with the affine pieces of projective space) and local coordinates around a point $x$ are by definition functions which project to a basis of the cotangent space $\mathfrak m_x/\mathfrak m_x^2.$ The problem (that I am aware of at least) is that you don't have the nice topological properties of manifolds. So while you can make arguments formally analagous to those for manifolds, you cannot use a theorem to say that the two cases are the same (that is, there is no theorem saying that manifolds are the same as varieties over an algebraically closed field over characteristic $p>0$). Often this is done when computing things with differential forms on a smooth variety. For example, computing the canonical divisor of $\mathbb P^n$ makes explicit use of local coordinates and transformations of differential forms in exactly the same way as one would in the case of smooth manifolds.