Looking for 'elementary' approach that deals with Hom$_R(\oplus_{i \in I}M_i, N) \overset{\simeq}\to \Pi_{i \in I} \text{Hom}_R(M_i,N)$ I am trying to make this question as clear as possible. I will have to elaborate a bit though in order to do so. 
I am in a first semester linear Algebra course (although people of a higher semster in my University claim that we're being taught Algebra rather than linear Algebra). This week I am struggling with the following problem:


Current Problem Set (8th week): To show, for $(N,+,\cdot)$ being an R-Module  $$\text{Hom}_R(\oplus_{i \in I}M_i, N) \overset{\simeq}\to \Pi_{i \in I} \text{Hom}_R(M_i,N) \\ \hom_R(N, \Pi_{i \in I} M_i) \overset{\simeq}\to \Pi_{i \in I} \hom_R(N,M_i) $$

Which of course, as you can guess, I have no idea* on how to approach 

To the Problem: Exhausting google just a little bit, one does find many relevant links addressing the above problem, for example on this math blog (crazy project) here. However  in the authors answer the person uses the universal property of the direct product. I have never been introduced to such a property in my course yet, which brings me to the word elementary that I have used in the title.
There are also related questions (I am not going to link them) on this website, the answers are of the caliber "trivial, indeed, it is easy to see that, it is clear that" by which I don't mean to mock the answerers - in fact they are in agreement to what my Professor would say to  this exercise.

No idea*: I fibbed there a little. On the level of definition I believe to know several things:


*

*The direct sum $\oplus_{i \in I}M_i$ is a sub-module of the direct product $\prod_{i \in I}M_i$ and the direct sum is defined to be $m_i=0$ for almost all $ i \in I$ where $(m_i)_{i \in I}$ is a familiy of functions. 

*The direct product is nothing but a set $M= \prod_{i \in I} M_i$ with a component wise defined addition and scalar multiplication for $(m_i)_{i \in I} \in M$

*An element $\varphi \in \hom_R(\oplus_{i \in I}M_i,N)$ would be a homomorphism with scalars in $R$ that maps from $\oplus_{i \in I}M_i$ to $N$. So to make this a bit more precise: $$\varphi \in \hom_R(\oplus_{i \in I} M_i, N) \implies \varphi : \begin{cases} \oplus_{i \in I}M_i & \longrightarrow N \\ (m_i)_{i \in I} & \longmapsto \varphi((m_i)_{i \in I}) \end{cases} $$

*An element of the 'direct product over the index set $I$ of all homomorphisms with scalars in $R$ from $M_i$ to $N$' would be a 'family of mappings' of the like $$\omega_i: \begin{cases}M_i & \longrightarrow N \\ m_i & \mapsto \omega_i(m_i) \end{cases} $$

I am uncertain if all of the above is right. To quote my professor "all the exercises are trivial as soon as the definitions are clear". So I assume that I still lack understanding of the definition, because it feels hard for me to define an isomorphism as required in the exercise. 
If someone could explain to me the taught-process on how to obtain the isomorphism or link to me relevant literature where the matter is explained I would appreciate it. Please note that I want to work with the definitions as 'purely' as possible and not use many theorems (chances are high that I don't know them). 
 A: Suppose you have
$$
\varphi\colon\bigoplus_{i\in I}M_i\to N
$$
For any $i\in I$, you can consider the map $\varphi_i\colon M_i\to N$ defined by composing with the “canonical embedding” of $M_i$ into the direct sum, let's call it $c_i$; thus $\varphi_i=\varphi\circ c_i$. Such morphisms define a family
$$
\check{\varphi}=(\varphi_i)_{i\in I}\in\prod_{i\in I}\operatorname{Hom}_R(M_i,N)
$$
Now, suppose you have
$$
f=(f_i)_{i\in I}\in\prod_{i\in I}\operatorname{Hom}_R(M_i,N)
$$
If $x=(x_i)_{i\in I}$ is an element in the direct sum, you can define
$$
\hat{f}(x)=\sum_{i\in I}f_i(x_i)\tag{*}
$$
Actually you should consider a finite subset $F\subseteq I$ such that $x_i=0$ whenever $i\notin F$ and set
$$
\hat{f}(x)=\sum_{i\in F}f_i(x_i)
$$
noting that this is independent of $F$ so long as it satisfies the condition above. This justifies the notation in (*).
It's just a matter of boring verifications to show that $\hat{f}$ is actually a morphism and that the two maps we built are inverse of one another.

The other situation is quite similar (and when you'll study some category theory you'll understand why). If
$$
\varphi\colon M\to\prod_{i\in I}N_i
$$
is a homomorphism, you can compose it with the “canonical projection”
$$p_i\colon\prod_{i\in I}N_i\to N_i$$
so getting a family
$$
\check{\varphi}\in\prod_{i\in I}\operatorname{Hom}_R(M,N_i)
$$
Conversely, given
$$
f=(f_i)_{i\in I}\in\prod_{i\in I}\operatorname{Hom}_R(M,N_i)
$$
you can define
$$
\hat{f}\colon M\to\prod_{i\in I}N_i
$$
by
$$
\hat{f}(x)=(f_i(x_i))_{i\in I}\in\prod_{i\in I}N_i
$$
Do the necessary verifications.
A: the first isomorphism is "slightly less trivial" than the second, which amounts to $f(n)_i \leftarrow\rightarrow f_i(n)$.
if $|I|$ is infinite then the direct sum of an $I$-indexed family of modules is a proper submodule of the direct product of the same family, so there is at least the appearance of something that requires proof. the fact that these two modules share the same homset is (waves hands) somewhat analogous to the compactness theorem in model theory.  what you wrote above captures the required idea.
