Question: "..can one compute the arithmetic genus of $C_m$ in terms of the arithmetic genus of $C$?"
Answer: You find an exerecise in Hartshorne, I.7.2 where for any "hypersurface" $Z(f) \subseteq \mathbb{P}^n_k$, the arithmetic genus
$$F1.\text{ }p_a(Z(f))=\binom{d-1}{n}$$
where $d:=deg(f)$. The hypersurface $H:=Z(f)$ is defined as the zero set of an irreducible degree $d$ polynomial $f(x_0,..,x_n)\in k[x_0,..,x_n]$ where $k$ is an algebraically closed field. The arithmetic genus is defined using the Hilbert polynomial $p_H(t)$:
$$p_a(H):=(-1)^{n-1}(p_H(0)-1).$$
Later in the book this invariant is generalized and shown to be independent of choice of embedding of $H$ into projective space. In HH.Ex.III.5.3 they define the arithmetic genus $p_a(X)$ for any projective scheme $X$ of dimension $r$ over a field as
$$F2.\text{ }p_a(X):=(-1)^r(\chi(\mathcal{O}_X)-1)$$
where $\chi(\mathcal{O}_X):=\sum_i (-1)^ih^i(X, \mathcal{O}_X).$ If formula $F1$ holds for non-irreducible hypersurfaces you get the formula
$$E1.\text{ }p_a(C_m)=\binom{dm-1}{2}=\frac{1}{2}(dm-1)(dm-2).$$
Note: The Euler charaxteristic $\chi(\mathcal{O}_X)$ can be calculated using Cech-cohomology. Hence you should calculate $H^i(C_m, \mathcal{O}_{C_m})$ using Theorem HH.III.4.5. If the calculation in HH.I.7.2 does not use the fact that $f(x,y,z)$ is an irreducible polynomial, it may be formula $E1$ holds for $C_m$.
Note: Let $F(x,y,z,t):=t^{dm}-f(x,y,z)^m \in k[x,y,z,t]_{dm}$. It follows $F$ is a homogeneous degree $dm$ polynomial in the variables $x,y,z,t$. Let $S:=Z(F) \subseteq \mathbb{P}^3_k$. There is a map
$$\pi: S \rightarrow \mathbb{P}^2_k$$
defined "in coordinates" by $\pi(x,y,z,t):=(x,y,z)$. The map $\pi$ is finite and if $k$ is algebraically closed it is surjective. You get an exact sequence
$$ \pi^*\Omega^1_{\mathbb{P}^2_k/k} \rightarrow \Omega^1_{S/k} \rightarrow \Omega^1_{S/\mathbb{P}^2_k} \rightarrow 0$$
and by HH.i.7.2 it follows
$$p_a(S):=\binom{dm-1}{3}$$
in the case when $S$ is an irreducible hypersurface in $\mathbb{P}^3_k$. Hurwitz theorem (HH.IV.2) gives for any finite separable morphism of degree $n$ of curves $f: C \rightarrow C'$, a formula relating the arithmetic genus $p_a(C)$ and $p_a(C')$.
The formula says that
$$2p_a(C)-2=n(2p_a(C')-2)+deg(R)$$
where $R$ is the ramification divisor of $f$. You should look for a more general version of this theorem. The non-reduced curve $C_m$ is in the ramification locus of $\pi$. Note also that since $\mathbb{P}^2_k:=\mathbb{P}(V^*)$ where $V:=k\{e_0,e_1,e_2\}$ it follows
$$p_a(\mathbb{P}^2_k)=(-1)^2p_a(Spec(k))=1$$
by HH.III.8.4. This gives two approaches: One direct approach and one using the map $\pi$. Below you may find references:
https://mathoverflow.net/questions/71950/higher-dimensional-version-of-the-hurwitz-formula
Doing such a project you will need to study the relative cotangent bundle $\Omega^1_{S/\mathbb{P}^n_k}$ to give a Hurwitz formula for singular surfaces.
Example: If you take a look in Fulton ("Intersection Theory", Ex.18.3.9)
you will find that for any finite etale morphism $f: X^n \rightarrow Y^n$ of projective schemes over $k$ of dimension $n$, there is an equality
$$\chi(\mathcal{O}_X)=deg(f) \chi(\mathcal{O}_Y),$$
and you get the following formula:
$$Et.\text{ }p_a(X)=deg(f)p_a(Y)+(-1)^n(deg(f)-1).$$
The above map $\pi$ is note etale, hence the formula $Et$ does not apply.
Note (dec 2022) : This is also stated as an exercise in Hartshorne (Ex.III.4.7) in the chapter on cohomology. You must write down the Cech complex and calculate the dimensions explicitly. There is no condition on the base field or if the polynomial is irreducible.
