(1): Not necessarily.
We could take $a$ to be any irrational number in $(0,1)$ and $b=1-a$. In this case, for all $n>0$, $na+nb=n$ and $\{na\}+\{nb\}=1$.
If 1, $a$ and $b$ are linearly independent over $\Bbb Q$, then by the Weyl equidistribution criterion, the sequence of points $(\{na\},\{nb\})$ will be equidistributed in the unit square. This means that $\{na\}+\{nb\}$ will not be equidistributed in the interval $[0,2]$ but will have a triangular probability density function, with maximum density at 1.
On the other hand, if $a=b$ and $a$ is irrational, then $\{na\}+\{nb\}=2\{na\}$ will be equidistributed in $[0,2]$, because $\{na\}$ is equidistributed in $[0,1]$. Also,
it is possible for $\{na\}+\{nb\}$ to be equidistributed in a subinterval of $[0,2]$. For example, if $a=\sqrt{2}$, $b=-1/\sqrt{2}$, set $z:=\{nb\}$; then $na=-2nb$, so
$$\{na\}+\{nb\}=\{-2z\}+z=\left\{
\begin{array}{l}
1-z, \qquad z\in(0,\frac12],\\
2-z, \qquad z\in(\frac12,1].
\end{array}
\right.$$
Since $z$ is equidistributed in $[0,1]$, it follows that $\{na\}+\{nb\}$ is equidistributed in $[\frac12,\frac32]$.
(2): Both $\{\{na\}+\{nb\}\}$ and $\{n(a+b)\}$ are integer translates of $na+nb$ which are in $[0,1)$, so $\{\{na\}+\{nb\}\}=\{n(a+b)\}$. However, $\{na\}+\{nb\}$ and $\{n(a+b)\}$ may differ by 1. For example, if you take $a=1/\sqrt{2}$, $b=1/\sqrt{2}$ and $n=1$, then $\{na\}+\{nb\}=2/\sqrt{2}=\sqrt{2}$, but $\{n(a+b)\}=\{\sqrt{2}\}=\sqrt{2}-1$.