Suppose that an Abelian $p$-group $G$ can be written as a direct sum of countably many finite cyclic groups. Let $a_1,...,a_k$ be generators of $k$ distinct summands of this direct sum, and let these elements have orders $p^{n_1},...,p^{n_k}$ respectively. Then for what values of $m$ are $a_1,...,a_k$ linearly independent over $\mathbb{Z}_{p^m}$? I mean that in the sense that if $0\leq c_1,...,c_k<p^m$ and $c_1a_1+...c_ka_k=0$ then $c_1=...=c_k=0$
I’m particularly interested in the case when not all of the $n_i$’s are equal.
In general, if you have $G=\bigoplus_{i\in I} A_i$, and $x_j\in A_{i_j}$, $j=1,\ldots,k$, and the $i_j$ are pairwise distinct, then (using additive notation) $$\alpha_1x_1+\cdots+\alpha_kx_k = 0\iff \forall j(\alpha_jx_j=0).$$ This simply because that’s how the direct sum works.
In your case, the $a_i$ are in pairwise distinct summands, so we definitely have that $c_1a_1+\cdots +c_ka_k=0$ if and only if $c_ia_i=0$ for each $i=1,\ldots,k$. This occurs if and only if $p^{n_i}|c_i$ for $i=1,\ldots,k$.
In particular, if $m\gt n_i$ for some $i$, then setting $c_j=0$ for $j\neq i$ and $c_i=p^{n_i}\lt p^m$ gives $0=c_1a_1+\cdots+c_ka_k$, yet $c_i\neq 0$. Thus, we must have $m\leq n_i$ for all $i$; that is, $m\leq \min(n_1,\ldots,n_k)$.
Conversely, if $m\leq\min(n_1,\ldots,n_k$, and $c_1a_1+\cdots+c_ka_k=0$, then $p^{n_i}|c_i$ for each $i$; since $0\leq c_i\lt p^m\leq p^{n_i}$, this forces $c_i=0$ for all $i$.
So the values of $m$ that give you the desired implication are precisely $1\leq m\leq \min(n_1,\ldots,n_k)$.