$2^k$. Construction: $\mathcal F = \{1, 2, \ldots, 2^k\}$.

Define $A(x)$ be the set of scales of $x$ and let $d(x,y) = \lfloor \log_2|x-y|\rfloor$. Key claim: Suppose $\mathcal F$ is a nonempty set of real numbers. Then, \[ \sum_{x\in \mathcal F} 2^{-|A(x)|}\leq 1. \]We seek to prove this by induction on $|\mathcal F|$. Let $\chi(\mathcal F)$ denote the left hand side. $\mathcal F = \{x_1,\ldots, x_n\}$. Our goal is to remove some of the $x_i$'s to obtain a new $\mathcal F'$ satisfying $\chi(\mathcal F') \geq \chi(F)$. Indeed, whenever \[ d(x_{i-1}, x_i) > d(x_i, x_{i+1}) = \cdots = d(x_{j-1}, x_j) < d(x_j, x_{j+1}) \]by removing whichever of \[ 2^{-|A(x_i)|} + 2^{-|A(x_{i+2})|} + \cdots, 2^{-|A(x_{i+1})|} + 2^{-|A(x_{i+3})|} + \cdots \](i.e. remove either $\{x_m\colon i\leq m\leq j, m\equiv i\pmod 2\}$ or $\{x_m\colon i\leq m \leq j, m\not\equiv i\pmod 2\}$) is smaller, we obtain a new set $\mathcal F'$ such that $\chi(\mathcal F') \geq \mathcal F$, which completes the induction. (Here, we let $x_0, x_{n+1}$ lie at $-\infty, +\infty$, respectively, and do not include this in the definition of $A$.)

Daniel Xu points out that one can motivate the key lemma by trying to add elements to an equality case. Furthermore, there are lots of equality cases, so the best way to prove it is to act locally. This motivates induction. Then, looking at when we can remove $x_1$ tells us that we win whenever $d(x_1,x_2) < d(x_2,x_3)$. It's not hard to get the complete proof from here.

It is pretty immediate that when $k=1$ the answer is $2$. With a simple argument, we can get that when $k=2$ the answer is $4$. It is thus natural to guess that the answer is $2^k$, and the construction should then be clear: to obtain the construction for a given $k$, we can have two copies of constructions for $k-1$ put really far away from each other. However, this is not the only construction. I actually realized really late that $1,2,\ldots,2^k$ is also a construction, and this construction is of a completely different flavor. In fact, there are a lot of other constructions. The intuition is that if we replace each element in a construction for $k=a$ with a small cluster of a construction for $k=b$, then this is actually a construction for $k=a+b$. I'm a bit lazy about explaining this clearly and it doesn't really matter... but the message is that there are many optimal constructions that do not seem to share a lot in common. However, there is one thing they share in common---every element in an optimal construction for $k$ always has exactly $k$ different value from it. This shows that maybe we should consider some quantity that is more global to capture this behavior. But what this quantity is is stilll vague at this point. One other observation to make is that when dealing with small $k$'s, the key ingredient is the following: if $x<y<z$ and $v(x,y)=v(x,z)=t$, then $v(y,z)<t$; also if $v(x,y)=v(y,z)=t$, then $v(x,z)=t+1$. A natural question to ask is the following: do we necessarily need to know exactly what the numbers are in the set $F$? Or does some weaker condition suffice? One natural guess is the following: suppose that the elements in $F$ are $x_1<x_2<\cdots<x_n$. Maybe it suffices to know $v(x_i,x_{i+1})$, and then we know that $v(x_i,x_{i+2})=\max(v(x_i,x_{i+1}),v(x_{i+1},x_{i+2}))+\{0 \textup{ or }1\}$, and when $v(x_i,x_{i+1})=v(x_{i+1},x_{i+2})$ we must pick $+1$. We can then maybe repeat this process and so on to compute arbitrary $v(x_i,x_j)$. This leads to the following idea. We let $t$ run through the possible values of $v(x_i,x_j)$ for $x_i\neq x_j\in F$, from the smallest to the largest. We also start with $n$ isolated vertex, each representing an element in $F$. Now in each round, we added edges whose endpoints have value $t$ from each other (i.e. the distance between the two endpoints lie in $[2^t, 2^{t+1})$). This more or less mimics the process above that computes the values. If for a vertex, there is an edge incident to it that is added in a given round, then we say that this vertex is activated this round. What we want to show is that if $n>2^k+1$, then there exists at least one vertex that is activated for $k+1$ different rounds. At this point, I am fairly convinced that I can solve this problem with enough effort. This is because there are actually a lot of constraints on how the edges can be added each round. For example, in the very first round, the add edges have to form one or several disjoint paths. As another example, in each round and for each vertex, its new neighbors have to be in at most two different cliques formed by the edges added prior to this round. Since those constraints are not used eventually, I don't want to list all of them. Feel free to play around with them, and I believe that you will soon be convinced that you have more clue in solving this problem now. However, to solve this problem completely this way, it seems that there has to be some clever potential function chosen. I didn't bother to think if this approach can work (I didn't have paper and pen with me at that point), so I decided to switch a direction---what if we start by considering the largest value instead of the smallest one? It is pretty immediate that $F$ can be split to three parts $A,B,C$ where $A$ is the set of elements whose value to the rightmost element is the largest possible value, $C$ is the set of elements whose value to the leftmost element is the largest posisble value, and $B$ is the rest of the elements. It is then natural to restrict ourselves on $A\cup B$ and $B\cup C$. The hope is that if every element in $F$ has at most $k$ different values, then every element in $A\cup B$ has at most $k-1$ different values among the set and same holds for $B\cup C$. If this were true, then we can conclude by induction. This is of course falst---what we can deduce is that every element in $A$ has at most $k-1$ different values among $A\cup B$. If we still want to proceed by induction, we have to deal with this asymmetry. Now recall the early observation that we might want to consider some global quantity. That observation is further confirmed by the asymmetry we have to deal with here. The last question to ask is---what quantity should we consider? Here is the intuition: the worst case for this strategy occurs when $|A|=|C|$, and we have $2|A|+|B|=|F|$. Also, in $A\cup B$, the elements in $A$ are only allowed to have $k-1$ different values, whereas the elements in $B$ are allowed to have $k$ different values. Therefore it is natural to think that elements that are only allowed to have $k-1$ different values are "twice as bad as" the elements that are only allowed to have $k$ different values, and then it is natural to consider the quantity $\sum_{x\in F}2^{-s(x)}$ where $s(x)$ is the numer of values $x$ have among the set $F$.

The answer is $2^k$. Construction: $1,2,\ldots,2^k$. Now to show that $2^k$ is the maximum possible, we prove the following strengthened statement. Lemma. Let $F$ be any finite set of real numbers, and for any $x\in F$ let $s_F(x)$ be the size of the set $\{v(x,y)|x\neq y\in F\}$. Then $\sum_{x\in F}2^{-s_F(x)}\leq 1$. Assuming the lemma, we know that if $s_F(x)\leq k$ for all $x\in F$, then $|F|\leq 2^k$, as desired. Proof of the lemmaWe prove by induction on $|F|$. It clearly holds when $|F|=1$. Now let $t=\max_{x,y\in F, x\neq y}v(x,y)$ and let $A=\{x\in F|\exists x<y\in F s.t. v(x,y)=t\}, C=\{x\in F|\exists y>x\in F s.t. v(x,y)=t\}$ and $B=F\backslash A\backslash C$. Clearly $A\cap C=\emptyset$ and $A,C$ are nonempty. Moreover, we have $v(x,y)\neq t$ for every $x\in A, x,y\in A\cup B$ . Therefore we have $s_{A\cup B}(x)\leq s_F(x)-1$ for all $x\in A$, and so by the induction hypothesis applied on $A\cup B$ we have $2\sum_{a\in A}2^{-s_F(a)}+\sum_{b\in B}2^{-s_F(b)}=1$. Similarly, we have $\sum_{b\in B}2^{-s_F(b)}+2\sum_{c\in C}2^{-s_F(c)}=1$. Adding up the two quantity, and then we can conclude by induction.