I think this answers your question, but I'm not entirely sure. Hopefully some bits of it are useful to you--let me know!
I think, first and foremost, you have a bit of confusion regarding the two uses of the word 'differential' here.
On one hand, when you speak of the differentials as $g=\dim H^0(X,\Omega^1_{X/\mathbb{C}})$ you're speaking of algebraic differentials on $X$. It follows from the general philosophy of the GAGA principle (although there is a simpler proof in this special case) that if $X^\text{an}$ denotes the analylitification of $X$ (i.e. the Riemann surface associated to $X$), then $H^0(X,\Omega^1_{X/\mathbb{C}})$ coincides with $H^0(X^\text{an},\Omega^1_{X^\text{an},\text{hol}})$. In other words, the algebraic differentials on $X$ correspond to the holomorphic differentials on $X^\text{an}$.
On the other hand, when you speak of de Rham cohomology, you are thinking of the smooth differentials on $X^\text{an}$ (thought of as just a smooth real manifold here). Indeed, the statement
$$H^i_\text{sing}(X^\text{an},\mathbb{R})\cong H^i_\text{dR}(X^\text{an})$$
(known as de Rham's Theorem) involves purely the smooth (real) manifold structure of $X^\text{an}$, not its complex structure.
Now, that's not to say that you can't compute the singular cohomology of $X^\text{an}$ from the holomorphic differentials, it's just more difficult. Namely, what is the proof of de Rham's theorem? Well, let's fix some smooth manifold $M$. Then,
$$0\to\underline{\mathbb{R}}\to\Omega^0_M\to\Omega^1_M\to\cdots\to\Omega^n_M\to 0$$
is a resolution of sheaves (here $\underline{\mathbb{R}}$ denotes the constant sheaf). But, since smooth manifolds have partitions of unity, any $C^\infty$-module ($C^\infty$ is the sheaf of smooth functions) is a so-called 'soft' sheaf, and thus acyclic. So, the $\Omega^i_M$'s (being $C^\infty$-modules) are acyclic. So, the above tells us that we can compute the (sheaf) cohomology of $\underline{\mathbb{R}}$ as the cohomology of the following chain complex (of abelian groups):
$$0\to\Omega^0_M(M)\to\Omega^1_M(M)\to\cdots\to\Omega^n_M(M)\to 0$$
But, this is precisely the de Rham cohomology of $M$ (by definition), and so we obtain the isomorphism
$$H^i(M,\underline{\mathbb{R}})\cong H^i_\text{dR}(M)$$
The, we can use the general fact that on a locally contractible space, (sheaf) cohomology of the constant sheaf agrees with singular cohomology to get
$$H^i_\text{sing}(M,\mathbb{R})\cong H^i(M,\underline{\mathbb{R}})\cong H^i_\text{dR}(M)$$
which is what we wanted.
You could try to run this argument through on some complex manifold $\mathcal{X}$, replacing smooth differentials with with holomorphic differentials, in a hope to prove something like $H^i_\text{sing}(\mathcal{X},\mathbb{C})$ is equal to the cohomology of the following chain of abelian groups
$$0\to\Omega^1_\text{hol}(\mathcal{X})\to\cdots\to\Omega^n_\text{hol}(\mathcal{X})\to 0\quad (*)$$
Of course, something is fishy here. I used $n$ to denote the index of termination both here, and in the sequence of differentials on the smooth manifold above. But, they're different $n$! If you view $\mathcal{X}$ as a smooth manifold, the 'n' that shows up in the smooth de Rham sequence is the real dimension of $\mathcal{X}$, and the $n$ that shows up in this holomorphic de Rham sequence is the complex dimension of $\mathcal{X}$. So, of course it can't be true that $H^i_\text{sing}(\mathcal{X},\mathbb{C})$ is the cohomology of this chain of abelian groups coming from holomorphic differentials, else the singular cohomology of a smooth manifold would be concentrated in degrees $0$ to $n$, of the (real) $2n$-dimensional smooth manifold $\mathcal{X}$. This certainly isn't the case in general (think of $\mathbb{P}^n$).
What goes wrong? Well, it's still true that
$$0\to\underline{\mathbb{C}}\to \Omega^1_\text{hol}\to\cdots\to\Omega^n_\text{hol}\to 0$$
is a resolution of sheaves. But, complex manifolds do NOT have holomorphic partitions of unity. So, there is no reason that the sheaves $\Omega^i_\text{hol}$ should be acyclic, and, in general, they aren't (think about $\mathbb{P}^1$ with $\Omega^1_\text{hol}=\mathcal{O}(-2)$). So, you can't compute
$$H^i_\text{sing}(\mathcal{X},\mathbb{C})\cong H^i(\mathcal{X},\underline{\mathbb{C}})$$
from the cohomology of chain of abelian groups $(\ast)$.
That said, there is a way you could try and fix this. You can try and compute sheaf cohomology from any resolution of a sheaf, but you have to be more clever if the resolution isn't acyclic. In particular, you need the fancy notion of hypercohomology of a complex of sheaves. Roughly, this takes your complex of sheaves $\mathcal{F}^\bullet$ (which could certainly not be acyclic) resolves each $\mathcal{F}^\bullet$ by acyclics $\mathcal{I}^{\mathcal{F},\bullet}$ and then takes the cohomology of the total complex $\text{Tot}(\mathcal{I}^{\mathcal{F},\bullet})$. In fancier language, this is just identifying your complex $\mathcal{F}^\bullet$ with its associated object in the derived category of sheaves of abelian groups on $X$. It then chooses within its derived class an equivalent (i.e. quasi-isomorphic) complex of acyclics, and computes the (normal) cohomology of that complex (the cohomology of its sequence of global sections).
The amazing thing is that if $\mathbb{H}^i(\mathcal{F}^\bullet)$ denotes the $i^{\text{th}}$ hypercohomology of the complex $\mathcal{F}^\bullet$, and $0\to\mathcal{F}\to\mathcal{F}^\bullet$ is some resolution, then
$$H^i(\mathcal{X},\mathcal{F})\cong \mathbb{H}^i(\mathcal{F}^\bullet)$$
even though the $\mathcal{F}^\bullet$ may not be a complex of acyclics.
So, what is true, is that while $H^i_\text{sing}(\mathcal{X},\mathbb{C})$ may not just be the cohomology of $(\ast)$, it is true that $H^i_\text{sing}(\mathcal{X},\mathbb{C})\cong \mathbb{H}^i(\Omega^\bullet_\text{hol})$, which is the closest one can get to a "holomorphic de Rham's theorem".
A truly amazing theorem of Grothendieck's (called the comparison theorem for algebraic de Rham cohomology) says that if now $X/\mathbb{C}$ is some smooth quasiprojective scheme (variety, if you'd prefer) then
$$\mathbb{H}^i(\Omega^i_X)\cong \mathbb{H}^i(\Omega^i_{X^\text{an},\text{hol}})$$
In words, the hypercohomology of the de Rham sequence of algebraic differentials is isomorphic the hypercohomology of the de Rham sequence of holomorphic differentials on $X^\text{an}$. But, by what we said above,
$$\mathbb{H}^i(\Omega^i_{X^\text{an},\text{hol}})\cong H^i_\text{sing}(X,\mathbb{C})$$
and so we see that
$$H^i_\text{sing}(X^\text{an},\mathbb{C})\cong \mathbb{H}^i(\Omega^i_X)$$
So, we can, in fact, compute the singular cohomology of the analytification of $X$ purely from the algebraic information of $X$'s algebraic differentials, it's just slightly more complicated.
Remark: I mentioned the comparison theorem here for general flavor, but, as you'll see below, you need much, much less in this specific instance of its use. In particular, we are implicitly proving a special case of this theorem below.
There are many other important/really nice applications of Grothendieck's comparison theorem (beyond the obvious reality check that algebraic de Rham cohomology does what we want), but that's best left to another post).
Now that I've said the difference in the various incarnations of the word 'differential' in your question, we can actually begin to answer "are these two theorems related". These two theorems being that the genus of a smooth projective curve $X/\mathbb{C}$ agrees with the topological genus of its associated Riemann surface $X^\text{an}$, and this holomorphic version of de Rham's theorem
The answer is, with this new understanding of differentials, yes! We can relate these two theorems.
There is something called the Hodge to de Rham spectral sequence that reads as follows
$$E^{p,q}_1=H^q(X,\Omega^p_{X^\text{an},\text{hol}})\implies \mathbb{H}^{p+q}(X,\Omega^\bullet_{X^\text{an},\text{hol}})$$
this is purely coming from the definition of hypercohomology, and the spectral sequence for a total complex.
Now, it's a deep theorem (morally equivalent to the Hodge decomposition) that this degenerates sequence degenerates on the first page. But, in the case of curves it's much simpler. Namely, the first pages looks like the following
$$\begin{matrix}\vdots & \vdots & \vdots & \vdots & \\ 0 & 0 & 0 & 0 & \cdots\\ H^1(X^\text{an},\mathcal{O}_{X^\text{an}}) & \to & H^1(X^\text{an},\Omega^1_{X^\text{an}\text{hol}}) & 0 & \cdots\\ H^0(X^\text{an},\mathcal{O}_{X^\text{an}}) & \to & H^0(X^\text{an},\Omega^1_{X^\text{an},\text{hol}}) & 0 & \cdots\end{matrix}$$
But, $H^0(X^\text{an},\mathcal{O}_{X^\text{an}})\to H^0(X,\Omega^1_{X^\text{an},\text{hol}})$ is just differentiation of constants (so zero), and $H^1(X^\text{an},\Omega^1_{X^{\text{an}},\text{hol}})\to H^1(X^\text{an},\mathcal{O}_{X^\text{an}})$ is just the Serre dual to this map, and so also zero. So, we do have degeneration on the first page.
Thus, we see that
$$H^1_\text{sing}(X^\text{an},\mathbb{C})\cong \mathbb{H}^1(\Omega^\bullet_{X^\text{an},\text{hol}})=H^1(X^\text{an},\mathcal{O}_{X^\text{an}})\oplus H^0(X^\text{an},\Omega^1_{X^\text{an},\text{hol}})$$
and so, finally using the GAGA principle, to say that
$$H^1(X,\mathcal{O}_X)\cong H^1(X^\text{an},\mathcal{O}_{X^\text{an}})\qquad H^0(X,\Omega^1_X)\cong H^0(X^\text{an},\Omega^1_{X^\text{an},\text{hol}})$$
we derive the isomorphism
$$H^1_\text{sing}(X^\text{an},\mathbb{C})\cong H^1(X,\mathcal{O}_X)\oplus H^0(X,\Omega^1_X)\qquad(**)$$
Why is this good? Well, Serre duality tells us that since we're on a smooth curve,
$$H^1(X,\mathcal{O}_X)\cong H^0(X,\Omega^1_X)^\vee$$
Moreover, we know precisely what the first singular cohomology of $X^\text{an}$, it's just $\mathbb{C}^{2 g(X^\text{an})}$ (where $g(X^\text{an})$ denotes the topological genus). Thus, comparing dimensions in $(**)$ we see that
$$2g(X^\text{an})=\dim\left(H^1(X,\mathcal{O}_X)\oplus H^0(X,\Omega^1_X)\right)=2\dim H^0(X,\Omega^1_X)=2g(X)$$
where $g(X)$ denotes the genus of $X/\mathbb{C}$ as a smooth projective (algebraic) curve. So, we're done!
Follow up question
I'm not quite sure what you mean. The entirety of the above used Čech (equiv. derived) cohomology constantly. If you clarify what you mean by this, I may be able to say more.
Best Answer
The formula you gives concerns plane algebraic curves, but, in general a curve cannot embed in $\bf CP^2$ but in $\bf CP^3$.