Robustness of splitting a junta
The answer is “yes”. The proof is by contradiction.

For notational convenience, let us denote the first $n/2$ variables by $x$ and the second $n/2$ variables by $y$. Suppose that $f(x,y)$ is $\delta$-close to a function $f_1(x,y)$ which depends only on $k$ coordinates of $x$. Denote its influential coordinates by $T_1$. Similarly, suppose that $f(x,y)$ is $\delta$-close to a function $f_2(x,y)$ which depends only on $k$ coordinates of $y$. Denote its influential coordinates by $T_2$. We need to prove that $f$ is $4\delta$- close to a $2k$-junta $\tilde f(x,y)$.

Let us say that $(x_1,y_1) \sim (x_2,y_2)$ if $x_1$ and $x_2$ agree on all coordinates in $T_1$ and $y_1$ and $y_2$ agree on all coordinates in $T_2$. We choose uniformly at random a representative from each equivalence class. Let $(\bar x, \bar y)$ be the representative for the class of $(x,y)$. Define $\tilde f$ as follows:
$$\tilde f(x,y) = f(\bar x, \bar y).$$

It is obvious that $\tilde f$ is a $2k$-junta (it depends only on variables in $T_1 \cup T_2)$. We shall prove that it is at distance $4\delta$ from $f$ in expectation.

We want to prove that
$$\Pr_{\tilde f}(\Pr_{x,y}(\tilde f(x,y) \neq f(x,y))) = \Pr(f(\bar x, \bar y) \neq f(x,y)) \leq 4\delta,$$
where $x$ and $y$ are chosen uniformly at random.
Consider a random vector $\tilde x$ obtained from $x$ by keeping all bits in $T_1$ and randomly flipping all bits not in $T_1$, and a vector $\tilde y$ defined similarly. Note that
$$\Pr(\tilde f(x,y) \neq f(x,y)) = \Pr(f(\bar x, \bar y) \neq f(x,y))= \Pr(f(\tilde x, \tilde y) \neq f(x,y)).$$

We have,
$$\Pr(f(x,y) \neq f(\tilde x, y)) \leq  \Pr(f(x,y) \neq f_1(x, y)) + \Pr(f_1(x,y) \neq f_1(\tilde x, y)) + \Pr(f_1(\tilde x,y) \neq f(\tilde x, y)) \leq \delta + 0 + \delta = 2\delta.$$

Similarly, $\Pr(f(\tilde x,y) \neq f(\tilde x, \tilde y)) \leq 2\delta$. We have 
$$\Pr(f(\bar x, \bar y) \neq f(x,y))  \leq 4\delta.$$
QED

It easy to “derandomize” this proof. For every $(x,y)$, let $\tilde f(x,y) = 1$ if $f(x,y) = 1$ for most $(x',y')$ in the equivalence class of $(x,y)$, and $\tilde f(x,y) = 0$, otherwise.