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The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. Call such an $x$ nice.

Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $hk(2X)$. Then any nice $x\in[X,2X]$ is within distance $hk(2X)$ from some nice $x\in F$, hence the cardinality of $F$ satisfies $hk(2X)|F|\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-hk(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $$|G|\gg X/k(2X)\gg X/k(X).$$ Now $[x,x+hk(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.

The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. Call such an $x$ nice.

Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $hk(2X)$. Then any nice $x\in[X,2X]$ is within distance $hk(2X)$ from some $x\in F$, hence the cardinality of $F$ satisfies $hk(2X)|F|\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-hk(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $$|G|\gg X/k(2X)\gg X/k(X).$$ Now $[x,x+hk(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.

Call $x\in\mathbb{N}$ nice if $\lambda_f(n)$ changes sign in $[x,x+k(x)]$. By the quoted result applied for $h=1$, the number of nice $x\in[X,2X]$ is $\gg X$. Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $k(2X)$. Then any nice $x\in[X,2X]$ is within distance $k(2X)$ from some nice $x\in F$, hence the cardinality of $F$ satisfies $|F|k(2X)\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-k(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $|G|\gg X/k(2X)\gg X/k(X)$. Now $[x,x+k(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.

The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. Call such an $x$ nice. 

Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $hk(2X)$. Then any nice $x\in[X,2X]$ is within distance $hk(2X)$ from some nice $x\in F$, hence the cardinality of $F$ satisfies $hk(2X)|F|\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-hk(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $$|G|\gg X/k(2X)\gg X/k(X).$$ Now $[x,x+hk(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.

Call $x\in\mathbb{N}$ nice if $\lambda_f(n)$ changes sign in $[x,x+k(x)]$. By the quoted result applied for $h=1$, the number of nice $x\in[X,2X]$ is $\gg X$. Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $k(2X)$. Then any nice $x\in[X,2X]$ is within distance $k(2X)$ from some nice $x\in F$, hence the cardinality of $F$ satisfies $|F|k(2X)\gg X$, i.e., $|F|\gg X/k(2X)\gg X/k(X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-k(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $|G|\gg X/k(X)$. Now $[x,x+k(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\ll X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.

Call $x\in\mathbb{N}$ nice if $\lambda_f(n)$ changes sign in $[x,x+k(x)]$. By the quoted result applied for $h=1$, the number of nice $x\in[X,2X]$ is $\gg X$. Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $k(2X)$. Then any nice $x\in[X,2X]$ is within distance $k(2X)$ from some nice $x\in F$, hence the cardinality of $F$ satisfies $|F|k(2X)\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-k(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $|G|\gg X/k(2X)\gg X/k(X)$. Now $[x,x+k(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.