I'm far from understanding what a general tensor is, at least from the category-theoretic construction using the universal property of the tensor product of vector spaces.
But I (think I) know that a linear map $f:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ admits uncountably-many representations as an $m\times n$ matrix over $\mathbb{R}$, and exactly one representation after you fix a basis (for both spaces?).
On the other hand, if you start with an arbitrary $m\times n$ matrix over $\mathbb{R}$, this matrix can represent uncountably-many linear maps unless you specify a basis (correct?).
So my question is: a general $(m,n)$-tensor is a map from where to where, and can it always be represented as a grid-like array of numbers?
Yep. You can think of an $(m,n)$-tensor (on a vector space $V$) as a multilinear map
$$ T \colon \underbrace{V^{*} \times \dots \times V^{*}}_{m \text{ times}} \times \underbrace{V \times \dots \times V}_{n \text{ times}} \rightarrow \mathbb{F}. $$
If you choose a basis $\beta = (e_1,\dots,e_k)$ for $V$ and let $(e^1,\dots,e^k)$ denote the corresponding dual basis for $V^{*}$ then the map $T$ is determined uniquely by the scalars $T(e^{i_1},\dots,e^{i_m},e_{j_1},\dots,e_{j_n})$ where $1 \leq i_1,\dots,i_m,j_1,\dots,j_n \leq k$ which are called the components of the tensor $T$ with respect to the basis $\beta$. Different choices of bases will give you different multiarrays which represent the same tensor with respect to different bases and a given multiarray of scalars can represent different tensors with respect different bases.