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.
Dear Tim, on page 31 they consider a ring $A$ and two $A$- algebras defined by their structural ring morphisms $f:A\to B$ and $g:A\to C$. They then define the tensor product as a ring $D=B\otimes _A C$ and want to make it an $A$- algebra. For that they must define the structural morphism $A\to D$ and they claim that it is given by the formula $a \to f(a)\otimes g(a)$.This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.
PS: To prevent misunderstandings, let me add that Atiyah-MacDonald is, to my taste, the best mathematics book I have ever seen, all subjects considered.
Best Answer
Here is the comment to this book in author's web page:
Also you can check my web page for all possible math books errata (not completed yet).