The theorem is the following:
Let $R$ be a principle ideal domain, and let $M$ be a finitely generated $R$-module. Suppose that
$M \cong R^s \bigoplus R/Ra_1 \bigoplus ... \bigoplus R/Ra_u$ (1)
$M \cong R^t \bigoplus R/Rb_1 \bigoplus ... \bigoplus R/Rb_v$ (2)
where the $a_i's, b_i's$ are non-zero, non-unit elements in $R$ s.t. $a_1 \mid a_2 \mid ... \mid a_u$ and $b_1 \mid b_2 \mid ... \mid b_v$.
Then $s = t, u = v$, and $Ra_i = Rb_i, 1 \leq i \leq u.$
In the proof, we have (by a theorem), that $(1) \implies M = F \bigoplus TorM, (2) \implies M = F' \bigoplus TorM$, where $F \cong R^s, F' \cong R^t$
This is where my question comes. They say that from this, we have that
$\frac{M}{TorM} = \frac{F \bigoplus TorM}{TorM} \cong F$
I am confused about where both the equality and the isomorphism came from. Thanks
It's nicer to think of things as coming from $\mathbb{Z}$, the 'easiest' PID to think about. If $M$ is a finitely generated $\mathbb{Z}$-module, then it is the sum of copies of some number of copies of $\mathbb{Z}$ and then the rest of the elements this misses, exactly the torsion elements which will each 'look like' $\mathbb{Z}/n\mathbb{Z}$ for some $n$. You can put the copies of $\mathbb{Z}$ together to get something like $\mathbb{Z}^m$ and the torsion elements together into one big set (which is still torsion because every element has finite order, not all necessarily the same), so some $\oplus \mathbb{Z}/n_i\mathbb{Z}$, which we will call Tor $M$, for the torsion elements. Then $M \cong \mathbb{Z}^m \oplus \text{Tor } M$.
For a PID that is not $\mathbb{Z}$, you have the same thing. So a number of copies of the ring (the free part analogous to the copies of $\mathbb{Z}$) $R^m$. Then you have the torsion part, which above is written $\oplus R/(a_i)= \text{Tor }M$. Then you have $M \cong R^m \oplus \text{Tor }M$.
As for the isomorphism, you have killed the torsion part in the quotient: $M/\text{Tor }M = (F \oplus \text{Tor })/\text{Tor }M \cong F$. The equality really could also be an isomorphism because you are thinking of $M$ not as 'itself' but instead in the form $F \oplus \text{Tor }M$, but this is just pedantic either way. Then quotienting by $\text{Tor }M$ kills all torsion only leaving you with the free part. If you need to, just work out the map: $F \oplus \text{Tor }M \to F$ sending $(f,f') \mapsto f$. Is the map a morphism? Is the map surjective? What is the kernel? Then what does the First Isomorphism Theorem say?