higher level view of construction of tensor product I am trying to to understand the construction of tensor product. Let $M,N,P$ be $R$-module ($R$ is commutative ring with unity). Consider the free $R$-module generated by $M \times N$ i.e.$ R^{(M \times N)}$. See the commutative digram below. $B$ denotes the submodule generated by the elements of the form $(a,b+c)- (a,c)-(a,b)$ and other relations. I am not getting how to proceed further. Provide an higher level view. I mean I need to prove that ker $f^{\ Tilda}$ is uniquely determined by $f$

 A: In the situation in your diagram, for $\tilde{f}$ to be uniquely determined, we want $f$ to be a bilnear map from $M \times N$ to $P$. This means that it is linear in each factor: $f(m + m',n) = f(m,n) + f(m',n)$, $f(m,n+n') = f(m,n) + f(m,n')$. and $f(rm,n) = rf(m,n) = f(m,nr)$, where $m,m' \in M$, $n,n' \in N$, and $r \in R$. The equalities expressed here are exactly the relations generating $B$, which is exactly what gives the tensor product this universal property. Let's see how we can use this to prove $\tilde{f}$ is uniquely determined.
Existence: Since $R^{M \times N}$ is free, we can define a morphism, call it $\phi$, to $P$ just by specifying where generators go. The natural thing to do is $(m,n) \mapsto f(m,n)$. To show this actually defines a map from $R^{M \times N}/B$, we need to show that $B$ is contained in the kernel. Again, this is a result of $f$ being bilinear: the generators of $B$ look like $(m+m',n) - (m,n) - (m',n)$, $(rm,n) - r(m,n)$, etc. The first of these is, by the definition of our map from $\phi$, sent to $f(m+m',n) - f(m,n) - f(m',n)$, but $f$ being bilinear tells us that this is 0. Similarly, the second generator here is sent to $f(rm,n) - rf(m,n)$, and again, $f$ being bilinear tells us that this is zero. Therefore $B$ is in the kernel. Therefore by the universal properties of quotients, we get a unique map $\tilde{f}:R^{M \times N}/B \rightarrow B$ such that $\tilde{f}p = \phi$, where $p$ is the quotient map $R^{M \times N} \rightarrow R^{M \times N}/B$. Note though, that we defined $\phi$ such that $f = \phi \iota$, where $\iota$ is the obvious map $M \times N \rightarrow R^{M \times N}$. Note also that, as you've labelled it, $\pi = p\iota$. Therefore $f = \tilde{f}p\iota = \tilde{f}\pi$, as needed.
Uniqueness: suppose $g$ is another map like $\tilde{f}$ making the diagram commute. Then, for all $(m,n) \in M \times N$, we must have $g(m,n) = g(\pi(m,n)) = f(m,n)$, but this was also the value taken by $\tilde{f}(m,n)$. Since the pairs $(m,n)$ generate $R^{M \times N}$, they generate $R^{M \times N}/B$, so since $\tilde{f}$ and $g$ agree on all these pairs, they agree on the whole module, and are the same map.
A: There are some pretty erudite ways to look at the construction $(V,W)\mapsto V\otimes W$, but I don't think they can ring a bell unless you come from category theory to actual concrete mathematics.
A safe bet between your desire of abstraction and an hands-on construction might be the universal property of the tensor product: there is an isomorphism
$$
\{\text{bilinear maps } V\times W \to P\} \leftrightarrows \hom_R(V\otimes W,P)
$$
which is induced by the map $V\times W \to V\otimes W$ that sends a pair $(v,w)$ into its equivalence class in the quotient $R^{|V\times W|}/relations$.
A nice exercise is to take any book listing properties of the tensor product and deduce the most you can (like for example the fact that there is an iso $V(WU)\cong (VW)U$, or the fact that $UV\cong VU$ where "nothing" between two modules means "$\otimes$") from the universal property alone.
A: A module is an Abelian group with a ring action.
Tensor product in nature is the composition of bimodules. 
Note that a left module ${}_AM$ over a commutative ring $A$ can be equally well be regarded as a right module, hence also as an ${}_AM_A$ bimodule.
If we forget the additive structure - i.e. we replace rings by monoids and modules by monoid actions, then the constuction of tensor product becomes easier:
Let $X, Y$ be sets, $X$ equipped with a right action and $Y$ with a left action of monoid $B$. Then their tensor product will be the set $X\times Y/\sim$ where $\sim$ is the equivalence relation generated by $\langle xr,y\rangle\sim\langle x,ry\rangle$.
If, in addition, a monoid $A$ acts on $X$ from left, compatibly with the right action of $B\ $  [i.e. $(ax)b=a(xb)$ holds], and $C$ acts on $Y$ from right, then $X\times Y/\sim$ becomes an $A$-$C$-biact.
Also, observe that your designated elements $r(a,b) - (ra,b)$ and $r(a,b) - (a,rb)$ will cause $r(a,b)=(ar,b)=(a,rb)$ in the quotient.
