A compact Riemann surface $X$ is in particular a compact real orientable surface. These surfaces are classified by their genus.
That genus is indeed the number of handles cited in popular literature; more technically it is
$$g(X)=\frac {1}{2}\operatorname {rank} H_1(X,\mathbb Z) = \frac {1}{2}\operatorname {dim} _\mathbb C H^1_{DR}(X,\mathbb C) $$ in terms of singular homology or De Rham cohomology.
Under the pressure of arithmetic, geometers have been spurred to consider the analogue of compact Riemann surfaces over fields $k$ different from $\mathbb C$: complete smooth algebraic curves.
These have a genus that must be calculated without topology.
The modern definition is (for algebraically closed fields) $$ g(X)=\operatorname {dim} _k H^1(X, \mathcal O_X)= \operatorname {dim} _kH^0(X, \Omega _X)$$
in terms of the sheaf cohomology of the structural sheaf or of the sheaf of differential forms of the curve $X$.
Of course this geometric genus is always $\geq 0$.
There is a more general notion of genus applicable to higher dimensional and/or non-irreducible varieties over non algebraically closed fields: the arithmetic genus defined by $$p_a(X)=(-1)^{dim X}(\chi(X,\mathcal O_X)-1)\quad {(ARITH)}$$ (where $\chi(X,\mathcal O_X)$ is the Euler-Poincaré characteristic of the structure sheaf).
[ Hirzebruch and Serre have, for very good reasons, advocated the modified definition $p'_a(X)=(-1)^{dim X}\chi(X,\mathcal O_X)$, which Hirzebruch used in his ground-breaking book and Serre in his foundational FAC]
For smooth projective curves over an algebraically closed field $g(X)=p_a(X)\geq 0$ : no problem.
It is only in more general situations that the arithmetic genus $p_a(X)$ may indeed be $\lt 0$
Edit
The simplest example of a reducible variety with negative arithmetic genus is the disjoint union $X=X_1\bigsqcup X_2$ of two copies $X_i$ of $\mathbb P^1$.
The formula $(ARITH)$ displayed above yields: $p_a(X)=1-\chi(X,\mathcal O_X)=1-(dim_\mathbb C H^0(X,\mathcal O_X)-dim_\mathbb C H^1(X,\mathcal O_X))=1-(2-0)$
so that $$p_a(X)=p_a(\mathbb P^1\bigsqcup \mathbb P^1)=-1\lt0$$
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 $2p_a(C) - 1 + C \cdot C$.
Best Answer
Let $K / k$ be finite extension of fields of degree $d$. Let $X$ be smooth geometrically connected projective curve over $K$ of genus $g$. The morphism $\mathrm{Spec} K \to \mathrm{Spec} k$ is projective (Exercise 3.3.22 of Liu's Algebraic geometry and arithmetic curves), therefore $X$ is projective over $k$. Then $$ p_{a,k}(X) = 1- \chi_k(X) = 1 - d \cdot \chi_K(X) = 1 - d (1-g). $$ You can take $g = 0$.
EDIT. More concretely, let $k$ be a field, let $f \in k[t]$ be a monic irreducible polynomial of degree $d$, let $\alpha \in \bar{k}$ be a root of $f$ and let $K = k(\alpha)$. If $f(t) = \sum_{i=0}^d a_i t^i$, then $\mathrm{Spec} K$ is isomorphic to $$ \mathrm{Proj} k[x_0,x_1] / (\sum_{i=0}^d a_i x_0^i x_1^{d-i} ). $$ Then $$ X = \mathbb{P}^1_K = \mathbb{P}^1_k \times_k \mathrm{Spec} K = \mathbb{P}^1_k \times_k \mathrm{Proj} k[x_0,x_1] / (\sum_{i=0}^d a_i x_0^i x_1^{d-i} ) $$ is a closed subscheme of $\mathbb{P}^1_k \times_k \mathbb{P}^1_k$, hence it is a closed subscheme of $\mathbb{P}^3_k$ by Segre embedding, i.e. $$ X = \mathrm{Proj} k[z_0, z_1, z_2, z_3] / (z_0 z_3 - z_1 z_2, \sum_{i=0}^d a_i z_0^i z_1^{d-i}, \sum_{i=0}^d a_i z_2^i z_3^{d-i}).$$