Example 4 Given the function $f(x)=\log _{2}(x+1)$, the graph of $y=f(x)$ is shifted 1 unit to the left, and then the y-coordinates of all points on the graph are stretched to twice their original length (the x-coordinates remain unchanged), resulting in the graph of the function $y=g(x)$. Then the maximum value of the function $F(x)=f(x)-g(x)$ is $\qquad$.
Given that $g(x)=2 \log _{2}(x+2)(x>-2)$, then
$$
\begin{aligned}
F(x) & =f(x)-g(x)=\log _{2}(x+1)-2 \log _{2}(x+2) \\
& =\log _{2} \frac{x+1}{(x+2)^{2}}=\log _{2} \frac{x+1}{x^{2}+4 x+4} \\
& =\log _{2} \frac{1}{\frac{x^{2}+4 x+4}{x+1}} \\
& =\log _{2} \frac{1}{x+1+\frac{1}{x+1}+2}(x>-1) .
\end{aligned}
$$

Since $x+1>0$, we have
$$
F(x) \leqslant \log _{2} \frac{1}{2 \sqrt{(x+1) \cdot \frac{1}{x+1}}+2}=\log _{2} \frac{1}{4}=-2,
$$

with equality holding if and only if $x+1=\frac{1}{x+1}$, i.e., $x=0$. Thus, $F(x)_{\text {max }}=F(0)=-2$.