The following definition is from p. 306 of the 2nd edition of John Lee's Introduction to Smooth Manifolds -- note that I have seen similar definitions elsewhere:
Let $V_1, \dots, V_k, W_1, \dots, W_l$ be real vector spaces, and suppose $F$ is a multilinear map from $V_1 \times \dots \times V_k$ to $\mathbb{R}$ and $G$ is a multilinear map from $W_1 \times \dots \times W_l$ to $\mathbb{R}$. Define a function $$F \otimes G: V_1 \times \dots \times V_k \times W_1 \times \dots \times W_l \to \mathbb{R} $$ $$F \otimes G (v_1, \dots, v_k, w_1, \dots, w_l)=F(v_1, \dots, v_k)G(w_1, \dots, w_l). $$ ... $F \otimes G$ is a multilinear map from $V_1 \times \dots \times V_k \times W_1 \times \dots \times W_l$ to $\mathbb{R}$ called the tensor product of $F$ and $G$.
Question: Why invent a special symbol $\otimes$ for the tensor product when it is just a pointwise product? Wouldn't it just be easier to write $FG$ or $F \cdot G$ or $F(\dots,v_i, \dots)G(\dots, w_j, \dots)$?
Seemingly being misled by the special symbol, I always misremember the tensor product as being something far more complicated than it actually is, and thus I never remember the actual definition until I read it again. So I was hoping that some motivation for the symbol will help me in the future to remember the definition for what it is, something simple, and to be less afraid of the object. That way I'll be able to use and do calculations with tensors much more readily, instead of panicking when I see the tensor product symbol and forgetting that it just denotes pointwise multiplication.
Let $F(x) := Ax$ and $G(y) := By$ for some real matrices $A$ and $B$.
Then $F\cdot G$ usually denotes the inner product map ($x^TA^TBy$) and
$FG$ usually denotes composition ($ABy$).
Now in this special case you have an example where your suggested altenatives are both occupied with meaning.
$F\otimes G$ is a linear map with matrix
$A \otimes B = \pmatrix{a_{11}B & a_{12} B & \cdots \\ a_{21} B & \ddots & \\ \vdots &&}$
in the canonical basis of $\mathbb R^n \otimes \mathbb R^m$ by the way.