The assumption that $|f_n(x)| \le M_n$ for every $x$ guarantees that each numerical series $\sum_n f_n(x)$ converges absolutely. We call the limit $f(x)$

To see that the convergence is uniform: let $\epsilon>0$ Then there exists $K$ such that $n>K$ implies $\sum_{n>K} M_n < \epsilon$ Now, if $k>K$

$$ |f(x)-\sum_{n=1}^k f_n(x)| = |\sum_{n>k} f_n(x) | \le \sum_{n>k} |f_n(x)| \le \sum_{n>k} M_n < \epsilon $$

The $\epsilon$ does not depend on $x$ so the convergence is uniform.