I'd finally like to know the exact way how it's done, if it's possible at all. I know sometimes it might be better to avoid euclid but I have to know how it's done by using this.
Here is an example I made myself and I hope it's a good one.
Find the multiplicative inverse of $11$ in $\mathbb{Z_{12}}$
We can also write: $\text{ gcd }(11,12)=1$
$$12=11\cdot1+1$$
$$11=1 \cdot 11+0$$
Now $1=12-11\cdot 1$
I don't think we can do anything from here now.. But what does this tell us, what's the multiplicative inverse now? How can we know it from this notation? Did I do it correct at all? Please do tell me!
Finding the multiplicative inverse of $a $ modulus $m $ is finding the solution to $ax \equiv 1 \mod m $
We know that equation has a solution iff $\gcd(a, m) $ divides 1, which implies $\gcd(a, m) = 1$
If that is the case, then from the euclidean algorithm to find the gcd you know you can have
$$1 = x_0a + y_0m $$
With $x_0, y_0 $ integers. Right? But taking that modulus $m $ we get
$$1 \equiv x_0a + y_0m \mod m \iff 1 \equiv x_0a \mod m$$
And thus $x_0$ is $a $'s multiplicative inverse.
For your specific case we have $\gcd(11, 12) = 1$ so 11 has a multiplicative inverse. From the euclidean algorithm we can see, as you wrote, that $1 = 1\cdot12 - 1\cdot11$.
Here we have $x_0 = -1$ and $y_0 = 1$. So it follows that $x_0 = -1$ is the multiplicative inverse of 11 modulus 12:
$$-1 \equiv 11 \mod 11 \Rightarrow (-1)\cdot11 \equiv 1 \mod 12 \iff (11)\cdot11 \equiv 1 \mod 12 \iff 121 \equiv 1 \mod 12$$
But $121 = 10\cdot12 + 1 \Rightarrow 121 \equiv 1 \mod12$ showing that 11 is its own multiplicative inverse.