There is a way to promote function space topologies so that convergence of nets behaves the way that one would expect it to behave. There are probably other ways of getting a topology for spaces of partial functions.
Suppose that $X,Y$ are topological spaces. Let $\mathcal{C}\subseteq P(X)$. Suppose that $A_{R}\subseteq Y^X$ whenever $R\in\mathcal{C}$. Give each space $A_R$ a topology $\mathcal{T}_R$, and suppose that the restriction mapping $j_{R,S}:A_R\rightarrow A_S$ is continuous whenever $R,S\in\mathcal{C},S\subseteq R$. Let $\mathcal{T}$ be the topology on $\bigcup_{R\in\mathcal{C}}A_R$ where $U\in\mathcal{T}$ precisely when $U\cap A_R$ is open in $A_R$ for each $R\subseteq X$ and whenever $S\subseteq R$, we have $j_{R,S}^{-1}[U\cap A_S]\subseteq U\cap A_R$. The canonical basis for the topology on $\bigcup_{R\in\mathcal{C}}A_R$ consists of the sets of the form $\bigcup_{R\supseteq S}j_{R,S}^{-1}[U]$ where $U$ is open in $A_S$. Therefore, the topology $\mathcal{T}_R$ is the subspace topology inherited from $\mathcal{T}$.
Proposition: A net $(f_d)_{d\in D}$ in the space $\bigcup_{R\subseteq X}A_R$ converges to a function $f$ if and only if there exists some $d\in D$ where
$\text{Dom}(f)\subseteq\text{Dom}(f_e)$ whenever $e\geq d$ and where the net
$(f_e|_{\text{Dom}(f)})_{e\in D,e\geq d}$ converges to $f$ in the topology $(A_\text{Dom}(f),\mathcal{T}_\text{Dom}(f))$.
Proof:
$\rightarrow$ Suppose that $(f_d)_{d\in D}\rightarrow f$ in $\bigcup_{R\in\mathcal{C}}A_R$. Let $U=\bigcup_{R\in\mathcal{C},R\supseteq\text{Dom}(f)}A_R$. Then there exists some $d\in D$ where if $e\geq d$, then $f_e\in U$. But this implies that if $e\geq d$, then $\text{Dom}(f_e)\supseteq\text{Dom}(f)$. Suppose now that $O\in\mathcal{T}_{\text{Dom}(f)}$ and $f\in O$. Then $\bigcup_{R\subseteq\text{Dom}(f)}j_{R,\text{Dom}(f)}^{-1}[O]\in\mathcal{T}$ and $f\in\bigcup_{R\subseteq\text{Dom}(f)}j_{R,\text{Dom}(f)}^{-1}[O]$. Therefore, since $(f_d)_{d\in D}\rightarrow f$, there exists some $d_1\in D$ with $d_1\geq d$ where $f_e\in\bigcup_{R\subseteq\text{Dom}(f)}j_{R,\text{Dom}(f)}^{-1}[O]$ whenever $e\geq d_1$. However, since $f_e\in\bigcup_{R\subseteq\text{Dom}(f)}j_{R,\text{Dom}(f)}^{-1}[O]$, we know that
$f_e|_{\text{Dom}(f)}=j_{\text{Dom}(f_e),\text{Dom}(f)}(f_e)\in O$. From this fact, we know that $(f_e|_{\text{Dom}(f)})_{e\geq d_1}$ converges to $f$ in the topology $(A_\text{Dom}(f),\mathcal{T}_\text{Dom}(f))$.
$\leftarrow.$ Suppose that $O\in\mathcal{T}$ and $f\in O$. Let $U=O\cap X_{\text{Dom}(f)}$. Then $U\in \mathcal{T}_{\text{Dom}(f)}$. Let $V=\bigcup_{R\supseteq\text{Dom}(f)}j_{R,\text{Dom}(f)}^{-1}[U]$. Then $V\in\mathcal{T}$ and $V\subseteq O$. Since there is some $d\in D$ where $\text{Dom}(f)\subseteq\text{Dom}(f_e)$ for $e\geq d$ and where $(f_e|_{\text{Dom}(f)})_{e\geq d}$ converges to $f$ in the topology $\mathcal{T}_{\text{Dom}(f)}$, we know that there is some $d_1\geq d$ where if $e\geq d_1$, then $f_e|_\text{Dom}(f)\in U$. But this implies that $f_e\in V\subseteq O$ whenever $e\geq d_1$. We conclude that $(f_d)_{d\in D}$ converges to $f$ in the space $\bigcup_{R\in\mathcal{C}}A_R$.
Q.E.D.
The topology $\mathcal{T}$ is not $T_1$ whenever there are $R,S\in\mathcal{C}$ with $R\subseteq S,R\neq S$. Let $\leq$ be the specialization ordering with respect to the topology $\mathcal{T}$. Then we observe that $f\geq f|_R$ whenever $R\in\mathcal{C},R\subseteq\text{Dom}(f)$. Since the specialization ordering is not the equality relation, we conclude that the topology $\mathcal{T}$ is not $T_1$.
If you are looking for a different topology on spaces of partial functions, one can always associate the partial functions with their graphs which are closed subsets of $X\times Y$, but we can give the collection of closed subsets of $X\times Y$ various topologies such as the Vietoris topology.
