I believe 3.) is actually easily fixed. Simply replace the bottom arrow with $\mathcal{C}\rightarrow\operatorname{Fun}(K\times\{0\}\amalg K\times\{1\},\mathcal{C}).$ On the other hand, I got seriously stuck for a few days on your issue #2. I agree with you that the proof as written does not work, but the result still holds with a different + substantially harder proof (I think you may also need to assume some condition on the cardinalities) - let me try to remember it. I don't immediately recall any thoughts on issue 1.)

My experience is that there are a huge number of mistakes in HTT (but remarkably, none of them seem to be essential), and proofs should be read as suggestions. Here are some comments about your specific questions:

Can you give a link to where Proposition 1 appears, to make it clearer what you are asking? As stated, it appears incorrect, unless by "Riemann surface" you really mean "algebraic curve, possibly singular" and you are picking a specific compactification.

If you require the scheme to be one-dimensional, then I think the answer should be yes, essentially because there will be a scheme parametrizing valuations on the underlying field of fractions... but I need to think a bit to make sure this can actually be made rigorous when you are not necessarily over a field.

No. Let $(S_0,p_0)$ be $(\mathbb{A}^2,(0,0))$. Iteratively define $S_{i+1}$ to be the blow up of $S_i$ along $p_i$ and $p_{i+1}$ to be an arbitrary point on the new exceptional divisor. You can glue all the $S_i-p_i$ into one scheme $S_{\infty}.$

@JonathanFrink Do you know of a written source?

I think you are misinterpreting the sentence in Arakawa-Frenkel. The result proved there leads to a construction of the "master chiral algebra", which indeed produces a functor from the LHS to the RHS. However, I don't think it is clear that this functor is an equivalence (but maybe this is my ignorance speaking). More generally, my understanding is that global quantum geometric Langlands is still a ways away from being a theorem.

@geometer: Yes, that is correct.

The usual Yoneda embedding is from $S^{\operatorname{op}}$ to $\mathcal{P}(S)$ - to get $j$, apply the Yoneda embedding to $S^{\operatorname{op}}.$

I don't see any reason for it to be compactly generated, even in a case as simple as $X$ a smooth projective curve, $n=1$, and $G=SL_2.$ There $\operatorname{Map}(X,BG)\equiv\operatorname{Bun}_G(X)$ is not quasi-compact, which you can maybe leverage to show that its category of quasicoherent sheaves is not compactly generated, maybe using techniques along the lines of section $12$ of arxiv.org/abs/1112.2402.

They are both special cases of the following: If the local expansion of $f(x_1,x_2,\cdots,x_n)$ at the origin is $f_1+f_2+\cdots f_d$, where $f_i$ has degree $i$, then the space of lines through the origin in $f=0$ is an intersection of the hypersurfaces corresponding to $f_1,f_2,\cdots, f_d$ in $\mathbb{P}^{d-1}.$ This follows from a short computation.