The smallest $c$ that the bound holds for is $c = \frac{1}{\sqrt 2 - 1} \approx 2.41$.
Lemmas 1 and 2 show that the bound holds for this $c$.
Lemma 3 shows that this bound is tight.
(In comparison, Juri's elegant probabilistic argument gives $c=4$.)
Let $c=\frac{1}{\sqrt 2 - 1}$.
Lemma 1 gives the upper bound for $k=0$.
Lemma 1:
If $f$ is $\epsilon_g$-near a function $g$ that has no influencing variables
in $S_2$, and $f$ is $\epsilon_h$-near a function $h$ that has no influencing variables
in $S_1$, then $f$ is $\epsilon$-near a constant function,
where $\epsilon \le \frac{(\epsilon_g+\epsilon_h)/2}{c}$.
Proof.
Let $\epsilon$ be the distance from $f$ to a constant function.
Suppose for contradiction that $\epsilon$ does not satisfy the claimed inequality.
Let $y=(x_1,x_2,\ldots,x_{n/2})$ and $z=(x_{n/2}+1,\ldots,x_n)$
and write $f$, $g$, and $h$ as $f(y,z)$, $g(y,z)$ and $h(y,z)$,
so $g(y,z)$ is independent of $z$ and $h(y,z)$ is independent of $y$.
(I find it helpful to visualize $f$ as the edge-labeling
of the complete bipartite graph with vertex sets $\{y\}$ and $\{z\}$,
where $g$ gives a vertex-labeling of $\{y\}$,
and $h$ gives a vertex-labeling of $\{z\}$.)
Let $g_0$ be the fraction of pairs $(y,z)$ such that $g(y,z) = 0$.
Let $g_1=1-g_0$ be the fraction of pairs such that $g(y,z) = 1$.
Likewise let $h_0$ be the fraction of pairs such that $h(y,z) = 0$,
and let $h_1$ be the fraction of pairs such that $h(y,z) = 1$.
Without loss of generality, assume that, for any pair such that $g(y,z) = h(y,z)$,
it also holds that $f(y,z) = g(y,z) = h(y,z)$. (Otherwise, toggling the value of
$f(y,z)$ allows us to decrease both $\epsilon_g$ and $\epsilon_h$ by $1/2^n$,
while decreasing the $\epsilon$ by at most $1/2^n$,
so the resulting function is still a counter-example.) Say any such pair is ``in agreement''.
The distance from $f$ to $g$ plus the distance from $f$ to $h$
is the fraction of $(x,y)$ pairs that are not in agreement.
That is, $\epsilon_g + \epsilon_h = g_0 h_1 + g_1 h_0$.
The distance from $f$ to the all-zero function is at most $1 - g_0 h_0$.
The distance from $f$ to the all-ones function is at most $1-g_1 h_1$.
Further, the distance from $f$ to the nearest constant function is at most $1/2$.
Thus, the ratio $\epsilon/(\epsilon_g+\epsilon_h)$ is at most
$$\frac{\min(1/2, 1-g_0 h_0, 1-g_1 h_1)}{g_0 h_1 + g_1 h_0},$$
where $g_0,h_0 \in [0,1]$ and $g_1 = 1-g_0$ and $h_1=1-h_0$.
By calculation, this ratio is at most
$\frac{1}{2(\sqrt 2 - 1)} = c/2$. QED
Lemma 2 extends Lemma 1 to general $k$ by arguing pointwise,
over every possible setting of the $2k$ influencing variables.
Recall that $c=\frac{1}{\sqrt 2 - 1}$.
Lemma 2: Fix any $k$.
If $f$ is $\epsilon_g$-near a function $g$ that has
$k$ influencing variables in $S_2$, and $f$ is $\epsilon_h$-near a function $h$ that
has $k$ influencing variables in $S_1$,
then $f$ is $\epsilon$-near a function $\hat f$
that has at most $2k$ influencing variables,
where $\epsilon \le \frac{(\epsilon_g+\epsilon_h)/2}{c}$.
Proof. Express $f$ as $f(a,y,b,z)$ where $(a,y)$ contains the variables in $S_1$
with $a$ containing those that influence $h$, while $(b,z)$ contains the
variables in $S_2$ with $b$ containing those influencing $g$.
So $g(a,y,b,z)$ is independent of $z$,
and $h(a,y,b,z)$ is independent of $y$.
For each fixed value of $a$ and $b$, define $F_{ab}(y,z) = f(a,y,b,z)$,
and define $G_{ab}$ and $H_{ab}$ similarly from $g$ and $h$ respectively.
Let $\epsilon^g_{ab}$ be the distance from $F_{ab}$ to $G_{ab}$
(restricted to $(y,z)$ pairs).
Likewise let $\epsilon^h_{ab}$ be the distance from $F_{ab}$ to $H_{ab}$.
By Lemma 1, there exists a constant $c_{ab}$ such that
the distance (call it $\epsilon_{ab}$)
from $F_{ab}$ to the constant function $c_{ab}$
is at most $(\epsilon^h_{ab} + \epsilon^g_{ab})/(2c)$.
Define $\hat f(a,y,b,z) = c_{ab}$.
Clearly $\hat f$ depends only on
$a$ and $b$ (and thus at most $k$ variables).
Let $\epsilon_{\hat f}$ be the average, over the $(a,b)$ pairs,
of the $\epsilon_{ab}$'s, so that
the distance from $f$ to $\hat f$ is $\epsilon_{\hat f}$.
Likewise, the distances from $f$ to $g$ and from $f$ to $h$
(that is, $\epsilon_g$ and $\epsilon_h)$ are the averages,
over the $(a,b)$ pairs, of, respectively, $\epsilon^g_{ab}$ and $\epsilon^h_{ab}$.
Since $\epsilon_{ab} \le (\epsilon^h_{ab} + \epsilon^g_{ab})/(2c)$
for all $a, b$, it follows that
$\epsilon_{\hat f} \le (\epsilon_g + \epsilon_h)/(2c)$. QED
Lemma 3 shows that the constant $c$ above is the best you can hope
for (even for $k=0$ and $\epsilon=0.5$).
Lemma 3: There exists $f$ such that $f$ is $(0.5/c)$-near two functions $g$ and $h$,
where $g$ has no influencing variables in $S_2$
and $h$ has no influencing variables in $S_1$,
and $f$ is $0.5$-far from every constant function.
Proof.
Let $y$ and $z$ be $x$ restricted to, respectively, $S_1$ and $S_2$.
That is, $y=(x_1,\ldots,x_{n/2})$ and $z=(x_{n/2+1},\ldots,x_n)$.
Identify each possible $y$ with a unique element of $[N]$,
where $N=2^{n/2}$.
Likewise, identify each possible $z$ with a unique element of $[N]$.
Thus, we think of $f$ as a function from $[N]\times[N]$ to $\{0,1\}$.
Define $f(y,z)$ to be 1 iff $\max(y,z) \ge \frac{1}{\sqrt 2}N$.
By calculation, the fraction of $f$'s values that are zero
is $(\frac{1}{\sqrt 2})^2 = \frac{1}{2}$,
so both constant functions have distance $\frac{1}{2}$ to $f$.
Define $g(y,z)$ to be 1 iff $y\ge \frac{1}{\sqrt 2}N$.
Then $g$ has no influencing variables in $S_2$.
The distance from $f$ to $g$ is the fraction of pairs $(y,z)$
such that $y<\frac{1}{\sqrt 2}N$ and $z\ge \frac{1}{\sqrt 2}N$.
By calculation, this is at most $\frac{1}{\sqrt 2}(1-\frac{1}{\sqrt2}) = 0.5/c$
Similarly, the distance from $f$ to $h$, where $h(y,z)=1$
iff $z\ge \frac{1}{\sqrt 2}N$, is at most $0.5/c$.
QED
