Standard Brownian motion is the process you describe: a continuous Gaussian process $B_t$ whose distribution at time $t$ is normal with mean zero and variance $t$ (or in higher dimensions, mean zero and covariance $tI$).
Some authors also use the term "Brownian motion" to refer to translations or scalings of this process; e.g. $B_t + x_0$, which at time $t$ has distribution $N(x_0,t)$, or $c B_t$, which has distribution $N(0, c^2 t)$. So adding the word "standard" serves to clarify when we are not doing that; when we really do mean the process with $N(0,t)$ distribution at time $t$ (i.e. $x_0 = 0$ and $c=1$).
There is no process analogous to Brownian motion that has distribution $N(0,1)$ at time $t$. For one thing, it would have to have either a random starting point, or a jump immediately after time 0, which are typically things we don't want to have in our definition of Brownian motion. (However, you might like to look up the stationary Ornstein-Uhlenbeck process, which is a nontrivial continuous process with random starting point whose distribution at every time $t$ is indeed $N(0,1)$.)
In other contexts, the term "Brownian motion" can refer to other processes that are somehow analogous to standard Euclidean Brownian motion. For instance, on a Riemannian manifold $M$, one can define a stochastic process whose generator is the Laplace-Beltrami operator; this process is often called "Brownian motion on $M$" because it plays that role.
Best Answer
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.