Given a sequence of random variables $\{X_n \}_{n=1}^\infty$, the first Borel-Cantelli lemma tells us that if there is a positive sequence $\{ a_m \}_{m=1}^\infty$ for which:
$$ a_m \overset{m\rightarrow\infty}{\longrightarrow} 0 \quad \text{and} \quad \sum\limits_{n,m=1}^\infty \mathbb{P}\big( \vert X_n\vert> a_m \big) <\infty \tag{$\circledast$} $$ Then $X_n$ converges almost surely to $0$. My question is whether there is also a converse relation, i.e, $X_n\rightarrow0$ almost surely implies that there exists a positive sequence $\{a_m \}$ such that $\circledast$ holds?
If $a_n \to 0$ and $\sum_n P(|x_n| >a_n) <\infty$ then $X_n \to 0$ almost surely, by Borel Cantelli. Assuming that this is the sum you intended I will give a counterexample for the converse. Consider $(0,1)$ with Lebesgue measure and let $X_n(\omega)=n$ for $0 <\omega <\frac 1 n$, $X_n(\omega)=0$ for $\omega \geq \frac 1 n$. Then $X_n \to 0$ at every point but $P(X_n >a_n)=\frac 1 n$ for so $\sum_n P(X_n >a_n)=\infty$ no matter sequence $(a_n)$ you choose.
Of course this example also works for your double sum.