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.
Best Answer
Suppose that we wanted to determine $M=\coprod_{i\in I}C_i$ from the category theory definition. Think about what we know:
$M$ is a module.
For each $i$, there is a morphism $\iota_i : C_i \rightarrow M$.
You can think of this defining an algebraic structure - just like the definition of a group or ring or module. So, we know that, if $c_1\in C_1$ and $c_2\in C_2$ then we know that $\iota_1 c_1 + 3\cdot \iota_2 c_2$ has to be in $M$, since it's a combination of elements we know must be in there. To be more explicit, the objects we are sure must be in the coproduct would have the form $$\alpha_{1}\iota_{i_1}c_{1}+\alpha_{2}\iota_{i_2}c_{2}+\ldots + \alpha_{k}\iota_{i_k}c_{k}$$ where the $\alpha_j$ come from the ring of coefficients and $c_j$ come from the corresponding $C_{i_j}$. We can always group any pair of terms with the same $\iota_{i_1}$ and could move the $\alpha_i$ inside of the $\iota$s to just get that the only element's we're sure must be in such a structure a of the form $$\iota_{i_1}c_1+\ldots + \iota_{i_k}c_k$$ for distinct $i_j$. While there are plenty of structures containing all this data (for instance: $M$ could be the zero module and all of these things could be equal!), the coproduct is the "freest" (initial) such structure, and hence contains only the elements that must be in $M$ and does not impose any extra relations between them. Since the set of expressions of the prior form do form a module under the right operations, we can find out that that is indeed the coproduct. Then, we can move to a less natural definition where we note that "finite sums of elements of $c_i$ under the inclusion maps" is easily isomorphic to "elements of the cartesian product $\prod_i C_i$ with only finitely many non-zero terms."
The product $M = \prod_i C_i$ invokes a different structure:
$M$ is a module.
For each $i$, there is a map $\pi_i : M\rightarrow C_i$.
This definition doesn't so much tell us about what the elements of $M$ are, but rather what we can do with them: for any element $m\in M$, we can extract an element of the cartesian product (of sets) $\prod_i C_i$ by applying each of the functions $\pi_i$ to $m$. The universal property says that $M$ is a terminal example of such an object, meaning that defining a map to $M$ is the same as defining a map to $\prod_i C_i$ such that each map to a coordinate is a morphism the corresponding to $C_i$. Of course, since this Cartesian product has the structure of a module where the morphisms into it are the same as the functions whose coordinates are each morphisms, this means that the product has to be $\prod_i C_i$.
Another, more formal way to deal with this is to note that in the category of $R$-modules, $$\operatorname{Hom}(R, M) \cong M$$ where we can regard the set of maps between two modules as a module by pointwise operations. Note that this is essentially considering the family maps $f_m(r)=r\cdot m$ for $m\in M$. The universal property essentially says $$\operatorname{Hom}(R, \prod_{i\in I}C_i) \cong \prod_{i\in I }\operatorname{Hom}(R, C_i) \cong \prod_{i \in I}C_i$$ where the later two products are products of sets, not modules - but where module structure can then be imposed in a natural way. This level of indirection is necessary since the universal property of a product specifies the maps out of a module, which doesn't inherently tell us much about the elements of that module - so we need to find a way to identify the elements of a module by knowing the maps out of it, and the relation $\operatorname{Hom}(R, M) \cong M$ encodes what we need to reason about elements.