The genus $g$ of a curve $C$ can be defined as follows:
If $C$ is non-singular, then $g=g(C)$ is the number that appears in the Riemann-Roch theorem.
If $C$ is singular, then we find a curve $C'$ that is non-singular, and birational to $C$ (you can find $C'$ by resolution of singularities, for example). In this case, $g(C)$ is defined to be $g(C')$.
In particular, a curve of genus $1$ does not need to be smooth. The fact that $g(C)=1$ only implies that $C$ is birational to a smooth curve $C'$ of genus $1$, and if $C'$ has a rational point, then $C'$ is an elliptic curve and it has a Weierstrass form. Since $C$ and $C'$ are birational, then $C$ is birational to a Weierstrass form (which still does not imply that $C$ is smooth).
I found the answers to this question quite useful here.
$\newcommand{\Spec}{\mathrm{Spec}}$
Scholze---> Katz-Mazur: I really wouldn't stress too much about this, to be honest. Probably Scholze should say that $p$ is locally of finite presentation and/or $S$ is locally Noetherian. Since the moduli spaces of such objects constructed is locally Noetherian, you really have no harm restricting to such a thing. Then, proper implies finite type and since S is locally Noetherian this implies that $p$ is locally of finite presentation. And then, yes, we use
[Tag01V8][1] If it makes you feel any better, his ultimate goal with this paper, and subsequent ones (which, incidentally, my thesis is a generalization of one of these papers) is to work in the same realm as the work of Harris-Taylor. In Harris-Taylor's seminal book/paper where they prove local Langlands for $\mathrm{GL}_n(F)$ they explicitly restrict only the schemes which are locally Noetherian (as does Kottwitz, if I recall correctly, in his original paper "On the points of some Shimura varieties over finite fields).
Katz-Mazur ---> Scholze: A smooth proper connected curve over a field is automatically projective. We may assume we're over $\overline{k}$. Let $X$ be a smooth proper conneced curve. Let $U$ be an affine open subscheme. Then, by taking a projectivization of $U$ (i.e. locally closed immerse $U$ into some $\mathbb{P}^n$ and take closure) and normalizations you can find an $X'$ which is smooth and projective containing $U$. Then, you get a birational map $X\dashrightarrow X'$. One can then use the valuative criterion to deduce this is an isomorphism.
An elliptic curve is connected. Note then that if $X/k$ is finite type, connected, and $X(k)\ne \varnothing$ then $X$ is automatically geometrically connected. Since any idempotents in $\mathcal{O}(X_{\overline{k}})$ must show up at some finite extension, it suffices to show that $X_L$ is connected for every finite extension $L/k$. Note that since $\Spec(L)\to \Spec(k)$ is flat and finite then same is true for $X_L\to X$, and thus $X_L\to X$ is clopen. Thus, if $C$ is a connected component of $X_L$ it's clopen (since $X_L$ is Noetherian) and thus its image under $X_L\to X$ is clopen, and thus all of $X$. Suppose that there exists another connected component $C'$ of $X_L$. Then, by what we just said the image of $C$ and $C'$ both contain any $x\in X(k)$. Note though that if $\pi:X_L\to X$ is our projection, then $\pi^{-1}(x)$ can be identified set theoretically as $\Spec(L\otimes_k k)=\Spec(L)$ and co consists of one point. This means that $C$ and $C'$, since they both hit $x$, have an intersection point. This is a contradiction. So an elliptic curve, being connected and having $E(k)\ne \varnothing$, is automatically geometrically connected.
Best Answer
Certainly the arithmetic genus depends on the scheme structure, and not just the underlying cycle; see the example in Hartshorne of two skew lines in $\mathbb P^3$ coming together in a flat family (and acquiring an embedded point) if you want an illustration of this. (Or just compute the arithmetic genus of the subscheme $XY = Y^2 = 0$ of $\mathbb P^2$, and compare it with arithmetic genus of the underlying reduced subscheme.)
In particular, if $C$ is not a codimension one cycle (a divisor), then I'm not sure that this question has that much of an answer, the reason being that arithmetic genus depends on having an actual subscheme of $X$ (or if you like, a point of the Hilbert scheme of the ambient variety $X$), whereas a cycle is less data than that.
If $X$ is a surface, so that $C$ is a Cartier divisor, then we can interpret $dC$ as being the $d$th power of this Cartier divisor, and hence get a well-defined subscheme, which varies in a flat family if $C$ does, and hence for which it makes sense to discuss the arithmetic genus. As Cantlog suggests in comments, the arithmetic genus will then be given by the adjunction formula. In particular, the self intersection $C\cdot C$ will play a role.
You can see this by comparing the case of $X$ being $\mathbb P^2$ vs. the case of $X$ being a smooth quadric surface (i.e. $\mathbb P^1 \times \mathbb P^1$). In the first case, if we take $C$ to be a line, then $2 C$ is the linear equivalence class of conics, whose arithmetic genus is again $0$. On the other hand, if we take $C$ to be one of the lines in one of the rulings of a quadric (i.e. $C = \mathbb P^1 \times \text{ a point }$), then $2C$ is linear equivalent to the disjoint union of two lines in the ruling, and has arithmetic genus equal to $-1$ (assuming I have the sign in the formula for arithmetic genus right).
The point is that the two lines crossing (with is another member of the linear equivalence class of $C$ in the first case) has one less point then two disjoint lines (the crossing point appears just once altogether instead of once on each line), thus $\chi(\mathcal O_C)$ is one less in the crossing case than in the disjoint case, thus $p_a := 1 - \chi(\mathcal O_C)$ is one more, i.e. is $0$ in the first case rather than $-1$ in the second case.
From a more formal point of view: the adjunction formula says that $2 p_a - 2 = K_X \cdot C + C \cdot C,$ and
this is not linear in $C$; e.g. you find that the arithmetic genus $p_a(2 C)$ is equal to $2p_a(C) - 1 + C \cdot C$.