Linear nets do not in general suffice.
Let $p$ be a free ultrafilter on $\omega$, and let $X=\{p\}\cup\omega$. Points of $\omega$ are isolated, and $U\subseteq X$ is a nbhd of $p$ iff $p\in U$ and $U\setminus\{p\}\in p$. A map $f:X\to X$ is continuous iff $f(p)=p$ and $f^{-1}[U]\in p$ for each $U\in p$.
Note that $p$ is not the limit of any sequence in $\omega$, so the only convergent sequences in $X$ are the trivial ones, those that are eventually constant. Now suppose that $\nu$ is a linear net in $\omega$ of uncountable cofinality. There must be some $n\in\omega$ that appears cofinally in $\nu$, which is therefore not eventually in the nbhd $X\setminus\{n\}$ of $p$. Thus, no linear net in $\omega$ converges to $p$, and the only convergent linear nets in $X$ are trivial. Since every function preserves the convergence of trivial nets, but not every bijection $f:X\to X$ preserving $p$ is continuous, linear nets do not suffice to characterize the continuous maps from $X$ to $X$.
In general let $\kappa=\sup\{\chi(x,X)^+:x\in X\}$, where $\chi(x,X)$ is the character of the point $x$ in $X$, i.e., the maximum of $\omega$ and the minimum cardinality of a local base at $x$. Then every point of $X$ has a local base of cardinality less than $\kappa$. Suppose that $f:X\to Y$ is not continuous. Then there are an $x\in X$ and an open nbhd $U$ of $f(x)$ in $Y$ such that for every open nbhd $V$ of $x$, $f[V]\nsubseteqq U$. Let $\mathscr{B}$ be a local base at $x$ of minimum cardinality. $\langle\mathscr{B}\supseteq\rangle$ is a directed set of cardinality less than $\kappa$. For each $B\in\mathscr{B}$ fix $x_B\in B\setminus f^{-1}[U]$; then $\langle x_B:B\in\mathscr{B}\rangle$ is a net in $X$ converging to $x$ such that $\langle f(x_B):B\in\mathscr{B}\rangle$ does not converge to $f(x)$, so $f$ is not continuous. Thus, if a function $f:X\to Y$ preserves limits of convergent nets of cardinality less than $\kappa$, $f$ is continuous.