I'm banging my head against Euclid's algorithm at the moment and I think I need some external input in order to gain a breakthrough....
Let's say we have a fraction like:
$$ \frac{216}{66}=3*66+18 $$
Given that $d|216,d|66$ I can accept intuitively that $d|18$. After all, if 216 and 66 are both multiples of $d$, removing a multiple of $d$ from a multiple of $d$ should just yield another multiple of $d$.
What I can't motivate is the other steps in the chain, that is to say why any subsequent terms should have the same greatest common divider as 216, 66 or 18.
Let's proceed to the next step.
$$\frac{66}{18}=3*18+12$$
Since 12 is a linear combination of 66 and 18, 66, 18 and 12 must share a greatest common divider, using the same reasoning like before. This greatest common divider (in my thinking) is however not guaranteed to be the same greatest common divider as the one between 216, 66 and 18.
In order to check this we must convert 12 into a linear combination of 216 and 18, since again, a multiple of d minus a multiple of must be another multiple of d.
$$ \frac{216}{66}=3*66+18 = 3*(3*18+12)+18=9*18+3*12=10*18+3*12 $$
Sadly, this does not tell us 12 is divisible by d. It only tells us the expression $3*12$ is divisible by d. In order for $3*12$ to be divisible by d either 12 OR 3 must be divisible by d. Since we can't tell if it's 12 or 3 that's divisible by d, any subsequent step in the algorithm becomes ambiguous....
$\frac{216}{66}\ne3*66+18$ but $$216=3\cdot66+18.$$ Hence if $d\mid216$ and $d\mid66,$ then $d\mid216-3\cdot66=18$ and conversely, if $d\mid66$ and $d\mid18$ then $d\mid3\cdot66+18=216.$ So, $$\gcd(216,66)=\gcd(66,18).$$
Similarly, $\frac{66}{18}\ne3*18+12$ but $$66=3\cdot18+12$$ hence $$\gcd(66,18)=\gcd(18,12).$$
Therefore, $\gcd(216,66)=\gcd(18,12).$ If you want to check that $12$ is a linear combination (with integral coefficients) of $216$ and $66,$ go backwards: $$12=66-3\cdot18=66-3\cdot(216-3\cdot66)=66(1+3\cdot3)-3\cdot216=66\cdot10-3\cdot216.$$
Edit: you wanted to "convert 12 into a linear combination of 216 and 18" but that is not part of Euclid's algorithm, and it is in fact impossible, since $12$ is not a multiple of $\gcd(216,18)=18.$