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(Note here that in order to answer Q2 you need to explain how we may think of $Y$ as the zero locus of a bunch of polynomials)

(Q3) (less interesting) Now that $Y$ is a $\mathbf{C}$-scheme, explain why the analytic map $f:Y\rightarrow X^{an}$ induces a map of $\mathbf{C}$-scheme $f:Y\rightarrow X$.

(Note here that in order to answer Q2 you need to explain how we may think of $Y$ as the being locally the zero locus of a bunch of polynomials So first my guess was that in order to get the existence of these polynomials one would have to use GAGA so this was the original set-up of my question. I was hoping that by using the definition of coherence one could try to define locally on analytic open sets of $\overline{Y}$ "enough meromorphic functions" on $\overline{Y}$ which then could be used to construct an embedding in a complex projective space of a suitable dimension. However in the answer that was suggested I don't see how this notion of coherence is used, it is kind of hidden and I just don't like that. At the end of the day one has to show that $Y$ may be viewed locally as the zero locus of polynomials).

(Q3) (less interesting) Now that $Y$ is a $\mathbf{C}$-scheme by (Q2), explain why the analytic map $f:Y\rightarrow X^{an}$ induces a map of $\mathbf{C}$-scheme $f:Y\rightarrow X$.

Added

So I'll try to rephrase the problem a little bit in order to focus on the part that I'm really interested in.

So let us assume that $X$ is a smooth affine variety over $\mathbf{C}$. So concretely one may think of $X=Spec(\mathbf{C}[x_1,\ldots,x_n]/(f_1,\ldots,f_r))$ where the $f_i$'s are polynomials in $n$ variables which satisfy a suitable Jacobian condition which expresses the fact that $X$ is smooth. So now suppose that $Y$ is a smooth connected analytic variety and that $f:Y\rightarrow X^{an}$ is a surjective finite unramified analytic cover of $X^{an}$.

(Q2) Is there a simple way to put a $\mathbf{C}$-scheme structure on $Y$ which is compatible with its analytic structure?

(Note here that in order to answer Q2 you need to explain how we may think of $Y$ as the zero locus of a bunch of polynomials)

(Q3) (less interesting) Now that $Y$ is a $\mathbf{C}$-scheme, explain why the analytic map $f:Y\rightarrow X^{an}$ induces a map of $\mathbf{C}$-scheme $f:Y\rightarrow X$.

(Q4) (this might be very easy to answer) The map of $\mathbf{C}$-scheme $f:Y\rightarrow X$ is quasi-finite. Is it necessarily finite, i.e., is $Y$ necessarily an affine $\mathbf{C}$-scheme?

Say that we solve (Q2) and that (Q4) is answered positively then we may think of $Y=Spec(\mathbf{C}[y_1,\ldots,y_m]/(g_1,\ldots,g_s)$ and from this description it is easy to see that you have "enough meromorphic functions on $Y$". For example take two distinct points $P,Q\in Y$ then we may always find a linear polynomial $l(y_1,\ldots,y_m)$ such that $l(P)=0$ and $l(Q)=1$.

Note that I mainly care about (Q2).

Let $X$ be a smooth quasi-projective variety (so irreducible) over $\mathbf{C}$. We may think of $X$ as a complex manifold which we denote by $X^{an}$. Of course the topology on $X^{an}$ is finer than the Zarisiki topology on $X$. Now let us suppose that we have a surjective finite unramified analytic cover

$f:Y\rightarrow X^{an}$.

Now for the sake of simplicity (I'm quite sure that one may relax considerably these assumptions) we will assume that there exists a normal projective variety $\overline{X}\supseteq X$ (as an open subset in the Z-topology) and that there exists a normal compact analytic variety $\overline{Y}\supseteq Y$ ( as an open subset in the analytic topology) and a finite ramified analytic covering map

$\overline{f}:\overline{Y}\rightarrow\overline{X}^{an}$

which extends the map $f$.

Then one may look at the analytic coherent sheaf $O_{\overline{Y}}$ push it forward by $f_{*}$ and obtain the following analytic coherent sheaf on $\overline{X}^{an}$:

$\mathcal{F}^{an}:=f_{*}{\mathcal{O}}_{\overline{Y}}$.

Now by GAGA we know that there exists a unique algebraic coherent sheaf $\mathcal{F}$ on $\overline{X}$ such that the

(1) The "analytification" of $\mathcal{F}$ is equal to $\mathcal{F}^{an}$.

By definition of coherence of $\mathcal{F}$ we know that

(2) For evey $x\in\overline{X}$ there exists a Zariski open set $U$ of $x$ such that the sequence of algebraic sheaves

$({O_{\overline{X}}|U})^n\ \rightarrow ({O_{\overline{X}}|U})^m\rightarrow\mathcal{F}|U\rightarrow 0$

is exact for some integers $m,n\in\mathbf{Z}_{\geq 0}$ (which may depend on $x$).

Now using $(1)$ and $(2)$ is there a simple way to deduce that $\overline{Y}$ is projective. Note that once we know that $\overline{Y}$ is projective then $\overline{Y}\backslash Y$ is analytically closed and therefore Zariski closed which implies that $Y$ is quasi-projective.

The conclusion that I was interested in was $Y$ is quasi-projective. So it seems that one may find a proof that $\overline{Y}$ is projective in Chap 12 of SGA1, but I'm sure that there must be a direct and easy way to deduce the algebraicity of $\overline{Y}$ using $(1)$ and $(2)$.

Let $X$ be a smooth quasi-projective variety (so irreducible) over $\mathbf{C}$. We may think of $X$ as a complex manifold which we denote by $X^{an}$. Of course the topology on $X^{an}$ is finer than the Zarisiki topology on $X$. Now let us suppose that we have a surjective finite unramified analytic cover

$f:Y\rightarrow X^{an}$.

Now for the sake of simplicity (I'm quite sure that one may relax considerably these assumptions) we will assume that there exists a normal projective variety $\overline{X}\supseteq X$ (as an open subset in the Z-topology) and that there exists a normal compact analytic variety $\overline{Y}\supseteq Y$ ( as an open subset in the analytic topology) and a finite ramified analytic covering map

$\overline{f}:\overline{Y}\rightarrow\overline{X}^{an}$

which extends the map $f$.

Then one may look at the analytic coherent sheaf $O_{\overline{Y}}$ push it forward by $f_{*}$ and obtain the following analytic coherent sheaf on $\overline{X}^{an}$:

$\mathcal{F}^{an}:=f_{*}{\mathcal{O}}_{\overline{Y}}$.

Now by GAGA we know that there exists a unique algebraic coherent sheaf $\mathcal{F}$ on $\overline{X}$ such that the

(1) The "analytification" of $\mathcal{F}$ is equal to $\mathcal{F}^{an}$.

By definition of coherence of $\mathcal{F}$ we know that

(2) For evey $x\in\overline{X}$ there exists a Zariski open set $U$ of $x$ such that the sequence of algebraic sheaves

$({O_{\overline{X}}|U})^n\ \rightarrow ({O_{\overline{X}}|U})^m\rightarrow\mathcal{F}|U\rightarrow 0$

is exact for some integers $m,n\in\mathbf{Z}_{\geq 0}$ (which may depend on $x$).

Q: Now using $(1)$ and $(2)$ is there a simple way to deduce that $\overline{Y}$ is projective?

Note that once we know that $\overline{Y}$ is projective then $\overline{Y}\backslash Y$ is analytically closed and therefore Zariski closed which implies that $Y$ is quasi-projective.

The conclusion that I was interested in was $Y$ is quasi-projective. So it seems that one may find a proof that $\overline{Y}$ is projective in Chap 12 of SGA1, but I'm sure that there must be a direct and easy way to deduce the algebraicity of $\overline{Y}$ using $(1)$ and $(2)$.