You can find the Arzelà–Ascoli in utmost generality (definitely more than you need) in Engelking's "General Topology", theorems 3.4.20 (p. 163) and 8.2.10 (p. 443). In your case, these theorems reduce to:
$F \subset C([0, \infty))$ is relatively compact in the topology of uniform convergence on compact subsets of $[0, \infty)$ if and only if $F$ is equicontinuous at every point $x \in [0, \infty)$ and $\{f(x) \mid f \in F\} \subseteq \mathbb R$ is bounded for every $x \in [0, \infty)$.
Minor nitpick: it is not clear if $N$ in your question is supposed to be natural; even if it were, you may easily replace it with real positive numbers, because every real number $x$ sits between its integer part $[x]$ and $[x]+1$, which are natural numbers. (In fact, $N$ plays the role of the interval $[0,N]$, which can be readily replaced with arbitrary compacts of $[0, \infty)$.)
Pick $x \in [0, \infty)$ arbitrary. In relation (1) take $N=x$. From the $\varepsilon - \delta$ definition of the concept of limit, taking $\varepsilon = 1$ there exists $\delta_1 > 0$ such that if $\delta < \delta_1$ then $\sup _{f \in F} V^x (f, \delta) \le 1$. Equivalently, $V^x (f, \delta) \le 1$ for all $f \in F$ and $\delta < \delta_1$. Explicitly,
$$\sup \{ |f(s) -f(t)| : 0 \le s, t \le x, \ |s-t| < \delta \} \le 1 \quad \forall f \in F \ .$$
In particular, taking $\delta = \frac {\delta_1} 2 < \delta_1$ in the above, we get that $|f(s) -f(t)| \le 1 $ for all pairs $0 \le s, t \le x$ with $|s-t| < \frac {\delta_1} 2$ and for all $f \in F$.
The interval $[0,x]$ can be covered with $n(x, \delta_1) := \left[ \frac {2x} {\delta_1} \right] + 1$ subintervals of length $\frac {\delta_1} 2$, whence (using the triangle inequality multiple times)
$$|f(x) - f(0)| \le \\
\le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) + f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) + \dots + f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\
\le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) \right| + \left| f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) \right| + \dots + \left| f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\
\le 1 + 1 + \dots + 1 = n(x, \delta_1) \ .$$
(Warning: my count above might be off by $\pm 1$ because of the endpoints, I'm never good at counting, but this doesn't change the end result: we have been able to find an upper bound for $|f(x) - f(0)|$ which is independent of $f \in F$.)
Finally, if $B = \{ |f(0)| \mid f \in F\}$ then
$$|f(x)| = |f(x) - f(0) + f(0)| \le |f(x) - f(0)| + |f(0)| \le n(x, \delta_1) + |f(0)| \in n(x, \delta_1) + B$$
and the right-hand side is obviously bounded, being the translate by the constant (with respect to $f \in F$) $n(x, \delta_1)$ of the bounded subset $B$.
(Notice that since $\{f(0) \mid f \in F\}$ was bounded, so will be $\{|f(0)| \mid f \in F\}$, trivially.)
Since $x$ was arbitrary, all the work above proves the pointwise boundedness of $F$.
The (uniform, but this is not needed) equicontinuity of $F$, on the other hand, comes practically for free, being encoded in (1), as you remark yourself in the question.
Since you have pointwise boundedness and equicontinuity, you have relative compactness in the topology of uniform convergence on compact subsets.
Substracting the continuous limit we can assume that $f_n(x)$ are decreasing with limit $0$. This easily implies the equicontinuity of $\{f_n:n\in\mathbb N\}$. Indeed, given $x_0\in K$ and $\varepsilon>0$ there is $n\in\mathbb N$ with $0\le f_n(x_0)\le \varepsilon/4$. The continuity of $f_n$ then gives a neighbourhood $U$ of $x_0$ with $|f_n(x)-f_n(x_0)|\le\varepsilon/4$ for all $x\in U$ and thus $0\le f_n(x) = f_n(x)-f_n(x_0)+ f_n(x_0)\le \varepsilon/2$. For all $m\ge n$ and $x\in U$, this implies $$|f_m(x)-f_m(x_0)|\le f_m(x)+f_m(x_0)\le f_n(x)+f_n(x_0)\le \varepsilon.$$ Using the continuity of $f_1,\ldots,f_{n-1}$ we can finally make $U$ smaller to obtain the same continuity estimate for all $m\in\mathbb N$.
Let us now assume that $f_n$ does not converge uniformly to $0$, i.e., $\|f_{n(k)}\|_K\ge \delta$ for some $\delta>0$ and some strictly increasing sequence $n(k)$ of integers. Arzelá-Ascoli then implies that another subsequence $(f_{n(k(\ell))})_{\ell\in\mathbb N}$ converges uniformly to some continuous $g$ with $\|g\|_k\ge\delta$. But since uniform convergence implies pointwise convergence, $f_n\to 0$ pointwise, and pointwise limits are unique we obtain the contradiction $g=0$.
The argument is quite similar to a possible proof of the Arzelá-Ascoli theorem (which quite often is somewhat hidden behind technicalities): Compactness in the topology of pointwise convergence is due to Tychonov's theorem and equicontinuity implies that this topology coincides with the much finer topology of uniform convergence on the given set.
Best Answer
As a starting point we need to take the correct version of Arzela-Ascoli for the compact-open (or in this case also uniform) topology: $S$ is relatively compact (the closure is compact) iff it is pointwise relatively compact, and equicontinuous.
In the weaker pointwise topology, we have more compact sets, and there the criterion is simply: $S$ is relatively compact iff it is pointwise relatively compact: this makes it a subset of a product space of compact intervals of the reals, and thus has a compact closure due to the Tychonoff theorem.
In both cases, to characterize compactness, add closed to the list of conditions.