If I understand your question correctly, you want the same sequence $(\alpha_m)$ for each sequence from the convergence field of your summation method.

I'll try to show that ultralimit can be realized as a summability method in the way you described.

Let $\mathcal F$ be any free ultrafilter. Let us define
$$D=\{(A,n); n\in A, A\in\mathcal F\}$$
and $$(A,n)\le (B,m) \Leftrightarrow A\supseteq B.$$
Then $(D,\le)$ is a directed set. For $\alpha=(A,n)\in D$ we define
$$c_{\alpha,k}=
\begin{cases}
1;&k=n,\\\\
0;&k\ne n.
\end{cases}$$
which means that $y_\alpha=x_n$.

Now $y_\alpha$ converges to $L$ if and only if $L$ is the $\mathcal F$-limit of the sequence $(x_n)$. (This is very similar to the usual correspondence between filters and nets in topological spaces, see e.g. Proposition 6.2. here.) This implies that every bounded sequence is summable.

However, if we choose any sequence $\alpha_m=(A_m,n_m)$ then the convergence of $(x_{\alpha_m})$ is in fact the convergence of $(x_{n_m})$. For any given sequence $(n_m)$ it is easy to exhibit an example of a bounded sequence $x$ such that $(x_{n_m})$ is not convergent.

The above example shows that your claim is not true for arbitrary directed sets. However, if you work with a directed set which contains a cofinal subset of order type $(\mathbb N,\le)$, then it is obviously true. (Which is the case in the example you suggested.) I suspect that it should work even for directed sets having any countable cofinal subset. There is a post at math.stackexchange related to the nets of this type: http://math.stackexchange.com/questions/41634/ordered-sets-that-are-like-sequences/