Prove that $\lfloor\frac{n+1}{2}\rfloor+\lfloor\frac{n+2}{4}\rfloor+\lfloor\frac{n+4}{8}\rfloor+\lfloor\frac{n+8}{16}\rfloor+ \dots=n$

The first term in your infinite sum (calling it $S(n)$), using the identity, is

$$\left[n\right] - \left[\frac{n}{2}\right]$$

The next term is

$$\left[\frac{n}{2}\right] - \left[\frac{n}{4}\right]$$

The next term is

$$\left[\frac{n}{4}\right] - \left[\frac{n}{8}\right]$$

And so on. So then the partial sum out to the $m$th term, $S(n,m)$, is just $[n] - [n/2^m]$. Taking this as $m$ approaches infinity:

$$\lim_{m \to \infty}S(n, m) = \lim_{m \to \infty}\left(\left[n\right] - \left[\frac{n}{2^m}\right]\right) = \left[n\right] = n.$$

For $x=\frac{n}{2^k}$, the identity you have means:

$$\left\lfloor \frac{n}{2^k}\right\rfloor =\left \lfloor \frac{n}{2^{k+1}}\right\rfloor+\left\lfloor \frac{n+2^k}{2^{k+1}}\right\rfloor$$

So prove by induction for any $m\geq 0$ that:

$$ n = \left\lfloor\frac{n}{2^m}\right\rfloor+\sum_{k=0}^{m-1}\left\lfloor\frac{x+2^{k}}{2^{k+1}}\right\rfloor$$

The pick $m$ so that $2^m>n$.

Let's start with the observation that, for any integers $a,b\gt0$,

$$\left\lfloor{a\over b}+{1\over2b}\right\rfloor=\left\lfloor{a\over b}\right\rfloor\quad(*)$$

Now define

$$f(n)=n-\left(\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\left\lfloor n+8\over16 \right\rfloor+\cdots \right)$$

Then

$$\begin{align} f(2n) &=2n-\left(\left\lfloor 2n+1\over2 \right\rfloor+\left\lfloor 2n+2\over4 \right\rfloor+\left\lfloor 2n+4\over8 \right\rfloor+\left\lfloor 2n+8\over16 \right\rfloor+\cdots \right)\\ &=2n-\left(\left\lfloor n+{1\over2} \right\rfloor+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\\ &=2n-\left(n+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\\ &=f(n) \end{align}$$

and

$$\begin{align} f(2n+1) &=2n+1-\left(\left\lfloor 2n+2\over2 \right\rfloor+\left\lfloor 2n+3\over4 \right\rfloor+\left\lfloor 2n+5\over8 \right\rfloor+\left\lfloor 2n+9\over16 \right\rfloor+\cdots \right)\\ &=2n+1-\left(\left\lfloor n+1 \right\rfloor+\left\lfloor {n+1\over2}+{1\over4} \right\rfloor+\left\lfloor {n+2\over4}+{1\over8} \right\rfloor+\left\lfloor {n+4\over8}+{1\over16} \right\rfloor+\cdots \right)\\ &=2n+1-\left(n+1+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\quad\text{using (*)}\\ &=f(n) \end{align}$$

Consequently $f(n)$ is a constant function for all $n\gt0$, so it suffices to evaluate

$$\begin{align} f(1)&=1-\left(\left\lfloor 1+1\over2 \right\rfloor+\left\lfloor 1+2\over4 \right\rfloor+\left\lfloor 1+4\over8 \right\rfloor+\left\lfloor 1+8\over16 \right\rfloor+\cdots \right)\\ &=1-(1+0+0+0+\cdots)\\ &=0 \end{align}$$