For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$.
But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ are not (for example, $E(X_2X_1)=0$ and $E(W_2W_1)=1$).
Brownian motions have the property of independent increments, meaning that for any disjoint intervals $[a, b]$ and $[c, d]$, $W(b) - W(a)$ is independent of $W(d) - W(c)$. However, it is not true that $W(s)$ and $W(t)$ are independent. Without loss of generality, suppose $t > s$. Then the distribution of $W(t)$ with the information $W(s) = k \neq 0$ is normal centered around $k$, not $0$ like the unconditioned distribution of $W(s)$. Since $W(s)$ and $W(t)$ are not independent, the variances cannot just be added to conclude it has variance $s + t$. To find the actual distribution of $W(s) + W(t)$, note that $W(t)$ can be written as the sum of independent increments of the Brownian motion:
$$W(t) = [W(t) - W(s)] + W(s) \implies W(t) + W(s) = [W(t) - W(s)] + 2 \cdot W(s)$$
However, note that $W(t) - W(s)$ and $W(s)$ describe two disjoint increments and thus we can now add their variances to obtain the actual distribution of $W(t) + W(s)$. It follows that
$$W(t) + W(s) \sim \mathcal{N}(0, t - s) + \mathcal{N}(0, 4s) = \mathcal{N}(0, t + 3s)$$
Note that if $s > t$, then $W(t) + W(s) \sim \mathcal{N}(0, s + 3t)$.
As for your second question, $W(2t + 2s) - W(2s)$ and $W(s + t) - W(s)$ may be independent, depending on whether they describe the increments of the Brownian motion in two disjoint intervals. The first expression describes the increments in the period $[2s, 2t + 2s]$ and the first one describes the increments in the period $[s, s + t]$. The intervals are disjoint if and only if $t < s$. So their sum is distributed according to $\mathcal{N}(0, 3t)$ if $t < s$. Note that if $t \geq s$ (in which case the two are not independent), we can again break the sum down into three independent increments of the Brownian motion: $[s, 2s]$, $[2s, s + t]$, $[s + t, 2t + 2s]$. Since the second interval is included in both $W(2t + 2s) - W(2s)$ and $W(s + t) - W(s)$, it follows that
$$[W(2t + 2s) - W(2s)] + [W(s + t) - W(s)] \sim \mathcal{N}(0, s) + 2\cdot\mathcal{N}(0, t - s) + \mathcal{N}(0, s + t)$$
where all the normal random variables on the right-hand side are independent. Hence,
$$[W(2t + 2s) - W(2s)] + [W(s + t) - W(s)] \sim \mathcal{N}(5t - 2s)$$
As a sanity check, the two answers should go to the same result when $t = s$, and indeed this is true.
Best Answer
The conditional distribution of $W_{t+h}$ conditionally on $\mathcal F_t$ is a random distribution, that is, a map $M:\Omega\to\mathcal M_1^+(\mathbb R,\mathcal B(\mathbb R))$, measurable with respect to $\mathcal F_t$, and such that, for every bounded measurable function $u$, $$ \mathrm E(u(W_{t+h})\mid\mathcal F_t)=\int_{\mathbb R} u\mathrm dM\quad\text{almost surely}. $$ The OP explains why $$ \mathrm E(u(W_{t+h})\mid\mathcal F_t)=\mathrm E(u(W_{t}+Z_h)\mid W_t)\quad\text{almost surely}, $$ where $Z_h$ is centered normal with variance $h$ and independent of $W_t$. Thus, for every bounded measurable function $u$, $$ \int_{\mathbb R} u\mathrm dM=\int_{\mathbb R} u(W_t+z)\mathrm d\gamma_h(z)\quad\text{almost surely}, $$ where $\gamma_h$ is the centered normal distribution with variance $h$. This proves that, for $\mathrm P$-almost every $\omega$, the distribution $M(\omega)$ is normal with mean $W_t(\omega)$ and variance $h$.
Now, my suggestion would be to forget what is written above and, instead, to find and read (and meditate) an excellent (and most congenial) presentation of this stuff (and much more) given in the little (but excellent) blue book called Probability with martingales by David Williams.