A couple of properties of normal distributions:
- the difference of two random variables $X_1 \sim N(\mu_1, \sigma^2_1)$ and $X_2 \sim N(\mu_2, \sigma^2_2)$:
$$ X_2 - X_1 \sim N(\mu_2 - \mu_1, \,\sigma^2_1 + \sigma^2_2)$$
- the variance of a random variable $X$ times a constant $c$ is
$$Var(c X) = c^2 \, Var(X)$$
Now, if $X_t \sim \sqrt{t} N(0, 1)$ is my random variable, I can compute $X_{t + \Delta t} - X_t$ using the first property above, as $$X_{t + \Delta t} - X_t \sim \sqrt{t + \Delta t} \, N(0, 1) - \sqrt{t} \, N(0, 1) = N(0, (\sqrt{t + \Delta t})^2 + (\sqrt{t})^2) = N(0, 2 t + \Delta t)$$
or instead, using the second property
$$\begin{split} X_{t + \Delta t} - X_t \sim &\sqrt{t + \Delta t} \, N(0, 1) - \sqrt{t} \, N(0, 1) =\\ &\left(\sqrt{t + \Delta t} - \sqrt{t}\right) N(0, 1) =\\ &N\left(0, (\sqrt{t + \Delta t} - \sqrt{t})^2\right) =\\ &N\left(0, \Delta t + 2 t \left(1 - \sqrt{1 + \frac{\Delta t}{t}}\right)\,\right) \end{split}$$
The second option should be the correct one, but why the first procedure is wrong, why it does not lead to the same result? Observing the outcomes, it is tempting to think that the first property is to be understood as an approximation.
Note it is NOT true that the sum or difference of two normal random variables is always normal. See here for a counterexample.
In the special case where two normal random variables $X\sim N(\mu_x,\sigma^2_x),Y\sim (\mu_y,\sigma^2_y)$ are independent, then they are jointly (bivariate) normal and then any linear combination of them is normal such that
$$aX+bY\sim N(a\mu_x+b\mu_y,a^2\sigma^2_x+b^2\sigma^2_y)\quad (1).$$
If $X_t=\sqrt t Z$, for $Z\sim N(0,1)$ it is clear that $X_t$ and $X_{t+\Delta t}$ are not independent so your first approach (i.e. using $(1)$) is invalid.