For simplicity, I'll just talk about varieties that are sitting in projective space or affine space. In algebraic geometry, you study varieties over a base field k. For our purposes, "over" just means that the variety is cut out by polynomials (affine) or homogeneous polynomials (projective) whose coefficients are in k.
Suppose that k is the complex numbers, C. Then affine spaces and projective spaces come with the complex topology, in addition to the Zariski topology that you'd normally give one. Then one can naturally give the points of a variety over C a topology inherited from the subspace topology. A little extra work (with the inverse function theorem and other analytic arguments) shows you that, if the variety is nonsingular, you have a nonsingular complex manifold. This shouldn't be too surprising. Morally, "algebraic varieties" are cut out of affine and projective spaces by polynomials, "manifolds" are cut out of other manifolds by smooth functions, and polynomials over C are smooth, and that's all that's going on.
In general, the converse is false: there are many complex manifolds that don't come from nonsingular algebraic varieties in this manner.
But in dimension 1, a miracle happens, and the converse is true: all compact dimension 1 complex manifolds are analytically isomorphic to the complex points of a nonsingular projective dimension-1 variety, endowed with the complex topology instead of the Zariski topology. "Riemann surfaces" are just another name for compact dimension 1 (dimension 2 over R) complex manifolds, and "curves" are just another name for projective dimension 1 varieties over any field, hence the theorem you described.
As for why Riemann surfaces are algebraic, Narasimhan's book explicitly constructs the polynomial that cuts out a Riemann surface, if you are curious.
When you say "deformation functor", you have to be careful to specify exactly which functor you are thinking about. There are two relevant deformation functors at play here.
One is the functor $D$ of abstract deformations of $Y$. If $A$ is an Artin local $k$-algebra, then $D(A)$ is the set of isomorphism classes of flat families $Y_A$ over $A$ whose central fiber is isomorphic to $Y$.
The other functor $D'$ is the functor of deformations of $Y$ inside $\mathbb{P}^n_k$. Here, $D'(A)$ is the set of isomorphism classes of flat families $Y_A$ together with an embedding $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$ such that the central fiber is $Y \rightarrow \mathbb{P}^n_k$ (not isomorphic to $Y$, but exactly $Y$, as we are talking about subschemes of $\mathbb{P}^n_k$).
The functor $D$ does not have any direct relationship to the Hilbert scheme, but the functor D' certainly does.
Given any scheme $X$ and a point $x \in X$, the local ring $\mathcal{O}_x$ defines a functor $F: (Art)_k \rightarrow (Sets)$ by setting $F(A) = Hom(Spec A, Spec \mathcal{O}_x)$. We say that $F$ is pro-represented by $\mathcal{O}_x$.
If you take the local ring $\mathcal{O}_{[Y]}$ of the Hilbert scheme (which you probably know to be representable by an honest scheme by a theorem of Grothendieck) at the point $[Y]$, then the functor it pro-represents is none other than $D'$.
Being pro-representable is stronger than having a versal deformation space; it means that the functor has a universal deformation space, which is the formal completion of the point whose local ring is doing the pro-representing.
So the (uni)versal deformation space of $D'$ is the formal completion of the point $[Y]$ in the Hilbert scheme. As for the versal deformation space of $D$, I am not really sure what it looks like, but there is a morphism $D' \rightarrow D$ of deformation functors given by forgetting the embedding of $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$.
In general, I think of a versal deformation space as an infinitesimal object; it is only keeping track of what happens when you deform $Y$ a little bit. The Hilbert scheme, on the other hand, is global; it is keeping track of all subschemes with the same Hilbert polynomial as $Y$, including ones which might be far away from $Y$ (if $Y$ was a point, for instance). Thus, you would expect the versal deformation space of the appropriate functor to map into the Hilbert scheme, rather than the other way around.
Best Answer
Yes, every algebraic differential form is holomorphic and yes, the differential preserves the algebraic differential forms. If you are interested in projective smooth varieties then every holomorphic differential form is automatically algebraic thanks to Serre's GAGA. This answers (3).
Concerning (1) and (2) I suggest that you consult some standard reference as Hartshorne's Algebraic Geometry.
Edit: As pointed out by Mariano in the comments below there are subtle points when one compares Kahler differentials and holomorphic differentials. I have to confess that I have not thought about them when I first posted my answer above.
Algebraic differential $1$-forms over a Zariski open set $U$ are elements of the module generated by $adb$ with $a$ and $b$ regular functions over $U$ (hence algebraic) by the relations $d(ab) = adb + b da$, $d (a + b) = da + db$ and $d \lambda =0$ for any complex number $\lambda$. Since these are the rules of calculus there is a natural map to the module of holomorphic $1$-forms over $U$. This map is injective, since the regular functions on $U$ are not very different from quotients of polynomials.
If instead of considering the ring $B$ of regular functions over $U$ one considers the ring $A$ of holomorphic functions over $U$ then one can still consider its $A$-module of Kahler differentials. If $U$ has sufficiently many holomorphic functions, for instance if $U$ is Stein, then one now has a surjective map to the holomorphic $1$-forms on $U$ which is no longer injective as pointed out by Georges Elencwajg in this other MO question.