The functor Over is covariant, and you do have that Over(X1∐X2)≅Over(X1)×Over(X2). However, I wouldn't say Over "turns coproducts into products", since that phrase usually
means that additionally the canonical maps Xi→X1∐X2 are sent to the canonical projections Over(X1)×Over(X2)→Over(Xi), and for that, of course, the functor would need to be contravariant.
In this example what is really going on is that Over(X) is equivalent to Fun(X,Set), the category of functors from X to Set (where the set X is regarded as a discrete category, i.e., the category whose objects are the elements of X and that only has identity morphisms). [The equivalence is given by sending f:A→X to the functor that sends an element x in X to its inverse image under f.] Of course, the functor Set→Cat given by Fun(_,Set) is contravariant, and it does turn coproducts into products in the sense I mentioned above. In summary, the covariant functor Over and the contravariant functor Fun(_, Set) agree on objects even though they have opposite variance, and the second turns coproducts into products.
Edit to answer the edit to the question: the pullback does indeed make Over(X) into a contravariant functor which is naturally isomorphic to Fun(_,Set). This is easy to check: if you have an object of Over(Y), say f:A→Y, and a function g:X→Y, then the inverse image of an element x in X under the pullback object of Over(X) is just the preimage of g(x) under f (the definition of pull-back is basically chosen to make this true).
A more natural problem would be to decide when the canonical morphism $\bigoplus_i A_i \to \prod_i A_i$ is an isomorphism. Clearly, this is the case if almost all $A_i$ are trivial. Notice that often when people say that $X,Y$ are not isomorphic, they really mean that a certain canonical morphism $X \to Y$ is not an isomorphism.
The question if there is any "random" isomorphism between $\bigoplus_i A_i$ and $\prod_i A_i$ is not that natural, because this isomorphism may just ignore the inclusions and the projections of the direct sum resp. direct product and therefore can be "wild".
Let us replace for the moment the category of abelian groups with the category of vector spaces over some fixed field $k$. Let $V_i$ be an infinite-dimensional vector space. Then it is known (MO/49551) that $\dim(\prod_i V_i)=\prod_i |V_i|$, where $|V_i|$ is the cardinality of (the underlying set of) $V_i$. Clearly, we also have $\dim(\oplus_i V_i) = \sum_i \dim(V_i)$. It is also well-known (math.SE/194281) that $|V_i|=\max(\dim(V_i),|k|)$. So the question if $\bigoplus_i V_i$ and $\prod_i V_i$ are isomorphic really only depends on the cardinal number equation $\sum_i d_i = \prod_i \max(d_i,q)$, where $q=|k|$ and $d_i=\dim(V_i)$. (The whole algebraic structure has disappeared!) Let us assume for the moment that $k$ is countable, i.e. $q \leq \aleph_0$. Then the equation becomes $\sum_i d_i = \prod_i d_i$. If $d:=d_i$ is constant and $|I|=\aleph_0$, the equation becomes $d = d^{\aleph_0}$. And this is perfectly possible, for example when $d=2^{\aleph_\alpha}$. Of course there are also examples when $d_i$ is not constant and where $I,k$ are arbitrary. (On the other hand, for some cardinals $d$, the equation $d=d^{\aleph_0}$ might as well be independent from ZFC!)
So you see, there are many families of infinite-dimensional vector spaces $V_i$ with $\bigoplus_i V_i \cong \prod_i V_i$, for example $V_i=\mathbb{R}$ over the base field $\mathbb{Q}$. In particular, their underlying abelian groups are isomorphic, contradicting the claim. (Notice that you cannot really write down this isomorphism, and therefore it is of no practical importance.)
Best Answer
We only have to use the definitons of the universal properties we want to prove.
(1) (Coproduct part) I suppose that you are assuming that the $A_i$ are disjoint, if not WLOG we can find a family of disjoint objects wich are isomorphic to each of the previous ones. Let $B$ be an object in the cathegory (a set), with a family of morphisms (maps) $\{f_i:A_i\rightarrow B\}$, we have to show now that there is a unique $f:C\rightarrow B$ such that for each $i$, $f\circ c_i=f_i$, i.e. making the corresponding diagram conmutative.
For existence, we define a function $f:C\rightarrow B$ as follows $$f(a)=f_i(a)$$ whenever $a\in A_i$. It is immediate to see that $f$ is well defined since the $A_i$ are disjoint, and that $f\circ c_i(a)=f_i(a)$ for all i and $a\in A_i$.
For uniqueness, suppose that there is other $g:C\rightarrow B$ with the same property; then since $g$ and $f$ have the same codomain and domain we only have to shown that $g(x)=f(x)$ for all $x\in C$. So let $x\in C$, then $x\in A_i$ only for one $i$ obtainig that $$f(x)=f_i(x)=g\circ c_i(x)=g(c_i(x))=g(x)$$ as desired.
Hence, the disjoint union of the $A_i$ is the desired coproduct in the cathegory $\mathcal{C}_1$.
(Product part) For the product, take an object in the desired cathegory (a set) $B$ with a family of mophisms $\{f_i:B\rightarrow A_i\}$. Then, we have to show that there is a unique $f:B\rightarrow P$ such that for each $i$, $p_i\circ f=f_i$, that as the previous case means making the corresponding diagram conmutative.
For existence of such an $f:B\rightarrow P$, define $$f(b)=(f_1(b),...,f_n(b))$$ for each $b\in B$. Then, with any problem we see well-defineness and it can be seen that for each $b\in B$ and $i$ we have $$p_i\circ f (b)=p_i(f(b))=p_i((f_1(b),...,f_n(b)))=f_i(b)$$ as desired.
For uniquenes, take $g:B\rightarrow P$ with the same property. Now, given a $b\in B$ for each $i$ we have that $$p_i\circ f (b)=f_i(b)=p_i\circ g(b)$$ So, the tuples $f(b)$ and $g(b)$ in the cartesian product have each of it componets equal and thus they must be equal. From this we have that for each $b\in B$, $f(b)=g(b)$, and since $f$ and $g$ share the codomain and domain they must be the same function.
Hence, the cartesian product of the $A_i$ is the desired product in the cathegory $\mathcal{C}_1$.
For (2) and (3) the previous proofs works perfectly, we have only to add the technical detail of shaw that $f:B\rightarrow P$ is in that case the corresponding linear map of homomorphism. Then, the rest of set theoric details continue the same.
EDIT: (I will add now the rest of the proofs.)
(2) First of all, we will notice that for the finite case of the family as noted by Arturo Magidin the direct sum of the $A_i$ and their cartesian product is the same -which in the infinite case is not true, thnaks that when the cartesian product the finiteness condition of tuples is useless-.
One, we have make this observation we can center in the cartesian product with vector space structure for our proof for similarity with other cases.
(Coproduct part) First, observe that the $c_i$ are linear maps. Once we have done that observation, we consider our object (in this case a vector space) $B$ with a family of morphisms (linear maps) $\{f_i:A_i\rightarrow B\}$. As before, we have to show that there is a unique morphism $f:\prod A_i\rightarrow B$ such that $f\circ c_i=f_i$.
For existence, consider $f: \prod A_i\rightarrow B$ given by $$f((a_1,...,a_n))=f_1(a_1)+...+f_n(a_n)$$ where $a_i\in A_i$. (See why this trick fails in general in non-abelian groups, and why is important conmutativity for it.) Then, we comprobe inmediately that for any $i$ and each $a_i\in A_i$ we have $$f\circ c_i(a_i)=f(c_i(a_i))=f((0,...,a_i,...,0))=f_1(0)+...+f_i(a_i)+...+f_n(0)=f_i(a_i)$$ as desired.
For uniqueness, consider a $g: \prod A_i\rightarrow B$ with he same property. Then, we hva for each $(a_1,...,a_n)\in\prod A_i$ that (by linearity) $g((a_1,...,a_n))=g(\sum (0,...,a_i,...,0))=\sum g((0,...,a_i,...,0))=\sum g(c_i(a_1,...,a_n))=\sum f_i(a_i)$ So, since $f((a_1,...,a_n))=\sum f_i(a_i)$ we obtain $f((a_1,...,a_n))=g((a_1,...,a_n))$. From where by cinciden of codomain and domain, $f=g$ as wanted.
(Product part) We only have to apply the result in the cathegory of sets, by showing that in the standard case of an object B and a family of morphisms $\{f_i:B\rightarrow A_i\}$ the map \begin{align} f:B&\rightarrow \prod A_i\newline b&\mapsto (f_1(b),...,f_n(b)) \end{align} is a linear transformation. Which is a basic result of linear algebra. Then we apply the previous set theoric conclusions thanks to the cathegory being concrete.
(3) The same a for vector spaces. The basic fact is that in that the candidate for product is the same that the one in the cathegory of sets, but adding some extra information to the set and that the candidate set theoric map \begin{align} f:B&\rightarrow \prod A_i\newline b&\mapsto (f_1(b),...,f_n(b)) \end{align} is a morphism -homomorphism- in this cathegory.