Let me add a few words referring in particular to the paper you cited: 
"From totally nonnegative matrices to quantum
matrices and back, via Poisson geometry" S. Launois and T.H. Lenagan

So we have a matrix 
$$( a \quad b )$$
 
$$ ( c \quad d ) $$

and the relations 

1) Four relations of the type XY = q YX 
 $$ ab = q ba, ac = q ca, bd = q ba , cd = q cd  $$
they cover four out of 6 relations between basic elements,
relations between diagonal terms a,d and anti-diagonal b,c are not covered.

2) relation for b,c is just commutativity: $$ bc = cb $$
 

3) and the last one is really "odd looking" here I will agree with you:

$$ ad - da = (q-q^{-1} ) bc $$


One way to motivate such relations is Manin's point of view - "coaction on a quantum place" - as explained by Theo Johnson-Freyd here:  https://mathoverflow.net/questions/204051/intuition-behind-the-definition-of-quantum-groups

Alternative point of view appeared before by Drinfeld:
Look at the Section 3.1 page 11 from the paper you cited: 
here you can see "classical limit" i.e. Poisson brackets for these relations :

1) Four relations of the type {X, Y}  = XY 
 $$ \{a, b \} =  ab, \{a, c \} =  ca, \{b, d \}=  ba , \{c, d \} =  cd  $$
they cover four out of 6 relations between basic elements,
relations between diagonal terms a,d and anti-diagonal b,c are not covered.

2) $$ \{ b, c \}  = cb $$
 

3) and the last one:
$$ \{a,d \} = 2bc $$ 

**The claim is that quantum relations above are quantization of the Poisson bracket relations here.** I.e. non-commutative algebra of observables is deformation quantization of the Poisson algebra here. 
Deformation quantization is now quite understood by efforts of Kontsevich et.al.
However without refering to big theories, as you a physicists you probably can easily see that
relations {X,Y} = XY should be quantized to XY = q YX - that are  relations of the type 1.
Why this is true: look at canonical quantization {p,q} = 1, quantized to [p,q]= h. 
Consider exponentials: {exp(p), exp(q) } = exp(p) exp(q) - we see exactly the same Poisson bracket as (1) - so it is just the exponents of canonical variables, so quantization is easy just exp(p) and exp(q) which in quantum world will give $$ exp(p) exp(q) = e^h exp(q) exp(p)$$,
so exactly the quantum relation (1) which we saw above.

{b,c} = 0, is quantized to bc = cb - is natutal

About the third relation - well I do not know easy way to quantize it from basic principles. If you know - tell me. 

So your question 
"Where do these odd looking relations come from? "
can be translated to the question: where do these Poisson relations come from ?

That is Drinfeld's insight.
He asked the following question - consider a Lie group and invariant Poisson bracket on it, we want Poisson structure would be COMPATIBLE with the group structure (in certain easy to guess sense). **And from that compatibility you will get that kind of Poisson relations.** Such groups are called Poisson-Lie groups as mentioned in answer by AHusain above. 

Concerning question 2:
In what sense are the matrices quantum (can the non-commutativity in the coordinate ring be understood as emerging from some quantisation procedure...)?

Yes as it was discussed above we have Poisson brackets which are quantized to that kind of noncommuative relations.

Question 3:
Should I think of quantum groups as a controlled mechanism for introducing non-commutativity into coordinate rings in general?

NO, that is not the case. 
Quantization needs only the Poisson bracket as an input.
Only in special case when Poisson bracket is somehow naturally related to some semisimple group you will get something related to quantum group.
In general you can take arbitrary Poisson bracket , quantize it and it will have nothing to do with the quantum groups.