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
I'm not an expert, the following is all just guesswork -- I similarly found the original papers unenlightening wrt their motivation.
As you said, the mystery mainly lies in the motivation of the additional step: modding out the functions from $X^{k+1} \to R$ by the subcomplex of functions which disappear on the neighborhood of the diagonal.
First, let's justify looking at neighborhoods of a space. We know from Alexander duality the philosophy of looking at tautness of a subspace $U$ with respect to a space $Y$.
We look at neighborhood $N$ of $U$ in Y (by neighborhood, we mean a subset $N$ of $Y$ that contains $U$ in its interior). The intersection of two neighborhoods of $U$ in $Y$ will be another neighborhood of $U$ in $Y$, so this gives us a system of groups $\{H^q(N)\}$ where $N$ ranges over all neighborhoods of $U$ in $Y$.
For each $N$, this gives us an inclusion $U \in N$, which induces a homomorphism $H^q(N) \to H^q(U)$. The subspace $U$ is said to be "tautly embedded" in $Y$ if this is an isomorphism for all $q$, all $N$, and all coefficient groups. Being taut implies that $U$ is compact and $Y$ is Hausdorff.
This gives us a hint: we are probably modding out by this subcomplex in order to deal with NON compact Hausdorff spaces.
Second, let's justify looking at the diagonal. The diagonal embedding $X \xrightarrow{\Delta} X \times X$, is simply a canonical way to embed a space X into an ambient space endowed with the product topology, $\Delta X := \{(x,x) \in X \times X\}$. It is useful when want to look in the neighborhood of a space $X$ (e.g., at germs of functions on $X$), but $X$ sits in no ambient space. The word, "diagonal embedding," comes from the example of embedding of $R^1 \hookrightarrow R^2$ taking $x \mapsto (x,x)$, that is, taking the line $R^1$ and embedding it into $R^2$ as the line $y=x$.
With this in mind, let's return our gaze to Alexander-Spanier cochains.
Here's my naive guess: modding out functions which disappear on any neighborhood of $X$, $N(X)$, artifically forces $X$ to satisfy the condition that $$H^q(\text{functions which disappear on }N(X)) \simeq H^q(\text{functions which disappear on }X)$$ for all $N$, all $q$, and all coefficient groups. Perhaps modding out by the subcomplex lets us "falsely" satisfy that $X$ is tautly embedded in $X \times X$, so that we may treat $X$ as if it were a compact space.
Below are a few additional comments toward why someone might have thought of modding out by that particular subcomplex.
Establishing notation: $X^{p+1}$ is the (p+1)-fold product of X with itself, that is, for $x_i \in X$, $(x_1, ..., x_{p+1}) \in X^{p+1}$.
$f^p(X) := \{$ functions $X^{p+1} \to \mathbb{Z} \}$, with functional addition as the group operation.
$f^p_0(X) :=$ elements of $f^p(X)$ which are zero in the neighborhood of the diagonal $\Delta X^{p+1}$
If we are examining functions defined pointwise on $X$, it’s natural to look at $X$-embedded in an ambient space, rather than the space $X$ itself. That is, $N(X)$ is the natural home of the jet bundle of $X$.
Functions which disappear on $N(X)$ form a group. If $f$ and $f’$ are both zero on $N(X)$ then $f-f’$ is zero on $N(X)$.
I'm not sure if the following is useful, nor how it fits into the story, but I figured I'd mention it.
The natural home of jet bundles (over a space $X$) is over the diagonal of X. From reading this paper, it seems that Grothendieck brought to the fore the kth neighborhood of the diagonal of a manifold $X$ when he was porting notions of differential geometry into algebraic geometry (this was then ported back into differential geometry by Spencer, Kumpera, and Malgrange). We'll use the standard notation $\Delta X \subseteq X_{(k)} \subseteq X \times X$. The only points of $X_{(k)}$ are the diagonal points $(x, x)$, but, we equip our space $X_{(k)}$ with a structure sheaf of functions, and treat $X_{(k)}$ as if it is made of "k-neighbor points" (x,y) where x and y are the closest points to one another, what Weil called "points proches").
To picture $X_{(1)}$, we might imagine $X$ with an infinitesimal normal bundle, for $X_{(2)}$, an infinitesimal bundle that’s ever so slightly larger of the second derivatives (as we need more local information to take the 2nd derivative), and so on.
If we think of a function $\omega: X_{(k)} \to R$ which vanishes on $X \subseteq X_{(k)}$ as a “differential k-form,” then maybe: