The Barr-Beck part of the proof is the following.
Theorem: Let $C$ be an essentially small abelian $k$-linear category and let $\omega : C \to \text{Vect}_f$ be an exact faithful functor from $C$ to the category of finite-dimensional $k$-vector spaces. Then this functor exhibits $C$ as the category of finite-dimensional comodules over a coalgebra $A$.
What we would like to do is to show 1) that $\omega$ is comonadic and 2) that the comonad is given by tensoring with a coalgebra $A$, so that coalgebras over the comonad correspond to $A$-comodules.
However, the difference between "comodule" and "finite-dimensional comodule" is extremely important here: it is in fact not possible to apply Barr-Beck directly to $\omega$ because $\omega$ in fact fails to have a right adjoint in general, as soon as $A$ is infinite-dimensional.
The fix is to pass to Ind categories on both sides, and so to consider, not $\omega$, but
$$\text{Ind}(\omega) : \text{Ind}(C) \to \text{Vect}.$$
These ind categories are locally presentable, so the locally presentable form of the adjoint functor theorem applies, and $\text{Ind}(\omega)$ has a right adjoint iff it preserves colimits. It preserves coequalizers because $\omega$ is exact, and it preserves filtered colimits by construction (I think; I haven't checked this carefully).
I am actually not sure how to complete the proof from here. I think one has to show that the right adjoint also preserves colimits, so that the corresponding comonad on $\text{Vect}$ preserves colimits. From here it is an exercise to show, e.g. using the Eilenberg-Watts theorem, that colimit-preserving comonads on $\text{Vect}$ are the same thing as coalgebras, and coalgebras over them are the same thing as comodules.
In the question it says $``\hat U_Y = $ $$\text{Spec} \varprojlim A/I^n = \varinjlim \text{Spec}\, A/I^n"$$ but this is not correct; these two ringed spaces are not equal, and even the underlying topological spaces are not equal. For example, take $A = k[x]$ and $I = (x)$. On the left hand side $$\text{Spec} \varprojlim\, k[x]/(x)^n = \text{Spec}\, k[[x]]$$ which is Spec of a DVR and has two points $(0), (x)$. On the right hand side, $\varinjlim\, \text{Spec}\, A/I^n$ is a colimit where each topological space has just a single point $(x) \in \text{Spec}\, k[x]/x^n$. In particular, the resulting ringed space is not a scheme. (Formal schemes usually aren't schemes, just ringed spaces.)
The same goes for the example in the comments, $\text{Spec}\, \mathbb Z_p$ is Spec of a DVR, has two points, while the formal spectrum of the point $\text{Spec}\, \mathbb F_p \hookrightarrow \text{Spec}\,\mathbb Z$ is a formal scheme with one point, and global sections equal to $\mathbb Z_p$ (thus not a scheme).
Besides not being equal to Spec of their global sections at the point-set level, formal schemes also behave differently algebraically than schemes when passing to an open set. For an affine scheme with global section $f \in A$, the open set $D(f)$ has global sections $A_f$. This will not be the case for formal schemes, as the localization happens before completion. When $A = k[x,y], I = (x), f = y$ for instance, on the open set $D(y)$ you can have elements like $\sum (x/y)^n$ where the power $y$ in the denominator is unbounded. There is no such element in $k[y][[x]]_y$.
As for your first question, the formal neighborhood of a point will have a complete stalks, but when the dimension of the subscheme is higher the formal neighborhood will not be complete. Intuitively this is because the formal neighborhood construction only takes the completion in directions orthogonal to the subscheme you complete.
Going back to the $k[x,y], I=(x)$ example, the stalk is complete "with respect to $x$" but not with respect to $y$. In other words, the stalk would be $k[x,y]_{(x,y)}[[t]]/(t-x)$ which is not a complete local ring, the completion being $k[x,y]_{(x,y)}[[t,s]]/(t-x,y-s)$ .
Best Answer
In general, the pattern is that the category of geometric objects locally modeled on algebraic objects from some category $A$ is generated under certain colimits by $A^{op},$ which is the category of formal duals of objects of $A.$ The manifold case is unusual since every manifold is, so to speak, affine. So the analytic space case will be much more like schemes. I won’t try to answer the last question because you should try to hone in on a specific definition of analytic space you’re interested in. As you can start to see here, there are a lot of very different options once you get beyond $\mathbb C:$ https://ncatlab.org/nlab/show/Berkovich+space