What would be the difference between $$ \otimes_B $$ and $$ \otimes $$ both in the following context and in general?
Let A be a ring with $$ B \subset A $$ and M a B-Module. We can construct the induced module $$ Ind_B^A M = A \otimes_B M $$ as the quotient of a free abelian group with generators from A × M by relations
$$(a_1+a_2)×m=a_1×m−a_2×m,a×(m_1+m_2)=a×m_1−a×m_2,ab×m=a×bm $$
Why is $$ \otimes_B $$ different to $$\otimes$$ ?
Source: http://math.berkeley.edu/~serganov/math252/notes4.pdf
As Tobias said, when you see something like $M \otimes N$ with unadorned tensor product where $M$ and $N$ are modules, what's usually happening is that there is some standard fixed base ring that is being suppressed in the notation. For instance, rather than rings I generally work with algebras over some field $k$ - for me $\otimes$ is always $\otimes_k$, whereas all other tensor products need to state what ring they're over explicitly. It's important to know what ring you're working over because tensoring by a different base ring usually gets you a different answer! For instance, $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ is something different from $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. Anyway, it depends on the context you're working in.
However, reading through the notes you linked I actually can't see an example of that happening. Instead, what I suspect may be confusing you is that it's convention to use the unadorned tensor product for elements: so in your example, given $a \in A, m \in M$, we can consider the element in $A \otimes_B M$ that corresponds to the equivalence class of $a \times m$ in the free abelian group with respect to the appropriate relations for the tensor product. That is generally written $a \otimes m$. I don't think it's common for people to write $a \otimes_B m$ and the notes you linked don't - I guess because when you're working with elements you know what object they're from and hence what base ring the tensor product is respect to.