I claim that either of your properties (sequences or nets) is equivalent to having only finitely many non-isolated points.
Property $S$: Any pointwise null sequence in $C(K)$ has an almost disjoint subsequence.
Property $N$: Any pointwise null net in $C(K)$ has an almost disjoint subsequence.
Properties $S^+$, $N^+$ are similar to Properties $S,N$, except the conclusion is only assumed to hold for sequences (resp nets) of non-negative functions. It follows from our proofs below that $C(K)$ has property $S$ iff it has Property $S^+$, and it has property $N$ iff it has Property $N^+$.
Given a topological space $K$ and a subset $L$ of $K$, let $L'$ denote the set of points of $L$ which are not isolated relative to $L$. Define the transfinite derivatives by $$K^0=K,$$ $$K^{\xi+1}=(K^\xi)',$$ and if $\xi$ is a limit ordinal, $$K^\xi=\bigcap_{\zeta<\xi}K^\zeta.$$
Fact $1$: Note that any closed, infinite subset of a compact set has a non-isolated point.
Fact $2$: If $C(K)$ has property $N$, then for any normalized, pointwise null net $(f_\lambda)$ in $C(K)$ and any $b>1$, there exist $\alpha, \beta$ such that $\|f_\alpha+f_\beta\|<b$.
Claim: Let $K$ be a compact, Hausdorff space. Then $C(K)$ has Property $N$ iff $C(K)$ has propety $S$ iff $K''=\varnothing$.
We first prove that if $K''=\varnothing$, then $C(K)$ has property $N$, and therefore also property $S$. Let $(f_\alpha)_{\alpha\in D}$ be a pointwise null net in $C(K)$. For $f\in C(K)$ and $\delta>0$, let $G(f,\delta)=\{x\in K:|f(x)|\geqslant \delta\}$, which is closed. We note that for $\delta>0$ and $f\in C(K)$, if $\|f|_{K'}\|<\delta$, then $G(f,\delta)$ is a finite subset of $K\setminus K'$. In particular, any member of $G(f,\delta)$ is isolated, since it is not in $K'$. To see this, $\|f|_{K'}\|<\delta$ immediately implies that $G(f,\delta)\cap K'=\varnothing$ and $G(f,\delta)\subset K\setminus K'$. If $G(f,\delta)$ were infinite, then since any closed, infinite subset of a compact space has a non-isolated point, it would follow that $\varnothing \neq G(f,\delta)'\subset G(f,\delta)\cap K'=\varnothing$, a contradiction. We recursively select finite sets $G_1, G_2, \ldots\subset K\setminus K'$ and $\alpha_1,\alpha_2, \ldots$ such that, with $G_0=\varnothing$ and $G_i=G(f_{\alpha_i},1/i)$, $\|f_{\alpha_i}|_{K'\cup G_1\cup \ldots \cup G_{i-1}}\|<1/i$. Let us see how to compete these choices. Assume $G_{i-1}$ has been chosen and is finite (which is trivially satisfied if $i=1$). Since $K'\cup G_1\cup \ldots \cup G_{i-1}$ is finite, there exits $\alpha_i\geqslant \alpha_{i-1}$ such that $\|f_{\alpha_i}|_{K'\cup G_1\cup\ldots\cup G_{i-1}}\|<1/i$. By the remarks above, $G_i:=G(f_{\alpha_i},1/i)$ is finite, as desired, This completes the recursive construction. Note that for each $x\in K\setminus K'$, the function $1_{\{x\}}$ is continuous, since $\{x\}$ is clopen when $x$ is isolated. Therefore for each $i\in\mathbb{N}$, $$g_i := \sum_{x\in G_i\setminus \cup_{j=1}^{i-1}G_j} f_{\alpha_i}(x) 1_{\{x\}}$$ is continuous. Fix $x\in K$ and note that $$ |f_{\alpha_i}(x)-g_i(x)| \leqslant \left\{\begin{array}{ll} 0 & : x\in G_i\setminus \cup_{j=1}^{i-1} G_j \\ \|f_{\alpha_i}|_{G_1\cup \ldots \cup G_{i-1}}\| & : x\in G_i\cap \bigl(\cup_{j=1}^{i-1} G_j\bigr) \\ 1/i & : x\in K\setminus G_i.\end{array}\right. \leqslant 1/i$$ Therefore $\lim_i \|f_{\alpha_i}-g_i\|=0$.
Next we have a lemma: Let $K$ be a compact, Hausdorff space with $K''\neq\varnothing$. There exist sets $U_m, C_m, U_{m,n}, C_{m,n}$ such that for each $m,n\in \mathbb{N}$, each $U_m$ and each $U_{m,n}$ is open, each $C_m$ and each $C_{m,n}$ is closed and non-empty, $$C_{m,n}\subset U_{m,n}\subset C_m\subset U_m.$$ Moreover, if $k,l\in \mathbb{N}$ are such that $k\neq m$ and $l\neq n$, then $U_m\cap U_k=\varnothing$ and $U_{m,n}\cap U_{m,l}=\varnothing$.
We postpone the proof of the lemma. For each $m\in\mathbb{N}$, fix a continuous $F_m:K\to [0,1]$ such that $F_m|_{C_m}\equiv 1$ and $F_m|_{K\setminus U_m}\equiv 0$. For each $m,n\in \mathbb{N}$, fix a continuous $f_{m,n}:K\to [0,1]$ such that $f_{m,n}|_{C_{m,n}}\equiv 1$ and $f_{m,n}|_{K\setminus U_{m,n}}\equiv 0$.
For each $m\in\mathbb{N}$, define $$g_m=F_m+\sum_{k=1}^{m-1}f_{k,m-k}\geqslant 0.$$ We claim that $\|g_m\|= 1$, which follows from the fact that $\|F_m\|= 1$ and $\|f_{k,m-k}\|= 1$, and the functions in the sum are disjointly supported. The functions $f_{1,m-1}, \ldots, f_{m-1,1}$ are supported on $U_1, \ldots, U_{m-1}$, and $F_m$ is supported on $U_m$. We also claim that $(g_m)_{m=1}^\infty$ is pointwise null. Indeed, fix $x\in K$. Consider several cases.
Case $1$: $x\in K\setminus \cup_{m=1}^\infty U_m$. Then $F_m(x)=f_{m,n}(x)=0$ for all $m,n$, and $g_m(x)=0$ for all $m\in \mathbb{N}$.
Case $2$: $x\in U_m$ (for a necessarily unique $m$) and $x\in U_m\setminus \cup_{k=1}^\infty U_{m,k}$. Then $f_{m,n}(x)=0$ for all $n\in \mathbb{N}$ and $F_n(x)=0$ for all $n>m$. Therefore $g_n(x)=0$ for all $n>m$.
Case $3$: $x\in U_{m,n}$ (for a necessarily unique $m,n$). Then $F_l(x)=0$ for all $l>m$ and $f_{k,l-k}(x)=0$ whenever $l>m+n$, since either $k\neq m$, in which case $f_{k,r}(x)=0$ for any $r$, or $k=m$ and $l-k>n$, in which case $f_{k,l-k}(x)=f_{m,l-m}(x)=0$, since $l-m\neq n$. Therefore $g_l(x)=0$ whenever $l>m+n$.
Fix $m<n$ and pick $x\in C_{m,n-m}$. Then $$g_m(x)=F_m(x)+\sum_{i=1}^{m-1}f_{k,m-k}(x)=F_m(x)=1,$$ since $x\in U_m\setminus \cup_{k=1}^{m-1}U_k$. Moreover, $$g_n(x) = F_n(x)+\sum_{k=1}^{n-1}f_{k,n-k}(x) = f_{m,n-m}(x)=1.$$ Therefore $\|g_m+g_n\|\geqslant 2$. This proves that $(g_m)_{m=1}^\infty$ can have no almost disjoint subsequence.
For the proof of the lemma, we require the following. In a compact, Hausdorff space $L$, for $x\in V\subset L$ with $V$ open, there exists an open set $W$ such that $x\in W\subset \overline{W}\subset V$. The proof of this is that since compact, Hausdorff spaces are normal, there exist disjoint open sets $W,B$ such that $\{x\}\subset W$ and $L\setminus V\subset B$. Then since $W\subset L\setminus B$ and the latter is closed, $\overline{W}\subset L\setminus B$. Since $B\subset L\setminus V$, $L\setminus B\subset V$.
We next claim that if $L\subset K$ is any set, $y\in L'$, and $U$ is an open neighborhood of $y$, then there exist sequences $(A_n)_{n=1}^\infty$, $(B_n)_{n=1}^\infty$, $(D_n)_{n=1}^\infty$, $(y_n)_{n=1}^\infty\subset L$ such that each $A_n,B_n,D_n$ is open, $D_1\supset D_2\supset \ldots$, each $D_n$ is a neighborhood of $y$, $A_n\cap D_n=\varnothing$, $A_n,D_n\subset U$, and $$y_n\in B_n\subset\overline{B_n}\subset A_n\subset V.$$ Indeed, since $y$ is not isolated in $L$, there exists $y_1\in L\cap U$ with $y_1\neq y$. There exist disjoint neighborhoods $A_1, D_1$ of $y_1$, $y$, respectively. By intersecting $A_1,D_1$ with $U$ if necessary, we can assume $A_1,D_1\subset U$. By the preceding paragraph, there exists an open set $B_1$ such that $y_1\in B_1$ and $\overline{B_1}\subset A_1$. Next, assume all choices have been made up to $n$. Since $U\supset D_1\supset \ldots \supset D_n$, there exists $y_{n+1}\in L\cap D_n$ with $y_{n+1}\neq y$. Choose disjoint open neighborhoods $A_{n+1}, D_{n+1}$ of $y_{n+1}$, $y$, respectively. By intersecting each with $D_n$ if necessary, we can assume $A_{n+1}, D_{n+1}\subset D_n$. Choose $B_{n+1}$ open with $y_{n+1}\in B_{n+1}$ and $\overline{B_{n+1}}\subset A_{n+1}$. This finishes this claim.
We now return to the proof of the lemma. Fix $x\in K''$. Apply the preceding paragraph with $x=y$, $L=K'$, and $U=K$ to get $A_n, B_n, D_n, y_n$. Let $U_n=A_n$ and $C_n=\overline{B_n}$. For each $n\in\mathbb{N}$, apply the preceding paragraph again with $x=y_n$, $L=K$, and $U=U_n$ to get new $A_m,B_m,D_m,y_m$. Let $U_{n,m}=A_m$, $C_{n,m}=\overline{B_m}$. Note that $y_m\in C_{n,m}\neq \varnothing$.
