Yes, I find it difficult to keep these things straight in my mind, too – in fact I thought I'd gotten them straight but found while writing this that they were less straight than I thought. :-)
I think a good starting point is to convince yourself that some of the operations involved commute. That is, if $a$ and $b$ are sequences of rational numbers and $[a]$ and $[b]$ denote their equivalence classes, then
$|[a]|=[|a|]$
and
$[a]-[b]=[a-b]\;.$
This doesn't work for $\le$, which instead obeys
$[a]\le [b] \Leftarrow\exists n_0\in\mathbb N:\forall n\gt n_0: a_n\le b_n\;.$
Now we can write the definition of the convergence of $[\{a_n\}]$ to $[a]$ (where $\{a_n\}$ is the constant sequence with elements $a_n$),
$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:|[\{a_n\}]-[a]|\le[\epsilon]$
(where I've written $[\epsilon]$ to make it easier to visually distinguish rationals and reals). Commuting the operations on the left-hand side of the inequality yields
$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:[|\{a_n\}-a|]\le[\epsilon]\;,$
and this is implied by
$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:\exists m_0\in\mathbb N:\forall m\gt m_0: (\{a_n\}-a)_m\le\epsilon_m\;.$
But $(\{a_n\}-a)_m=\{a_n\}_m-a_m=a_n-a_m$, so this becomes
$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:\exists m_0\in\mathbb N:\forall m\gt m_0: a_n-a_m\le\epsilon_m\;,$
which looks a lot more helpful, since the left-hand side of the inequality is now the difference of two rationals in the sequence, which converges since $a$ is Cauchy. We can simplify this to
$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall m,n\gt n_0:a_n-a_m\le\epsilon_m\;,$
since moving the existential quantifier in front of the universal one and requiring $m_0$ and $n_0$ to be the same only makes the condition stronger. Now it's recognizable that this condition is implied by $a$ being Cauchy, since $\epsilon_m$ must eventually be greater than some rational \epsilon'\gt0 and then we obtain $n_0$ from the Cauchy condition.