Denote the upper and lower Riemann integrals of $f$ by $\overline \int_I f$ and $\underline \int_I f$ respectively.
We have that $U(f+g,P) \le U(f,P) + U(g,P)$ and $L(f,P)+L(g,P)\le L(f+g,P)$. Thus, $$ L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P) \le U(f,P)+U(g,P). $$
Note that the right-hand inequality gives us that, for each partition $P$, $$ \overline \int_I (f+g) \le U(f,P) + U(g,P), $$ so that $$ \overline \int_I (f+g) \le \overline \int_I f + \overline \int_I g .$$ Similarly, $$ \underline\int_I f + \underline\int_I g \le \underline\int_I (f+g). $$
We know that $f$ and $g$ are Riemann-integrable, which means that their upper and lower Riemann integrals are the same. This allows us to conclude that $$ \int_I f + \int_I g = \overline\int_I (f+g) = \underline\int_I (f+g),$$ which is what we needed to show.
Under definition (1) or (2) we can show that a function $f$ cannot be both unbounded and Riemann integrable.
This can be shown by producing an $\epsilon > 0$ such that for any real number $A$, no matter how fine the partition, there is a Riemann sum with
$$|S(f,P) - A| > \epsilon$$
Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have
$$|S(f,P) - A| = \left|f(t_j)(x_j - x_{j-1}) + \sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right| \\ \geqslant |f(t_j)|(x_j - x_{j-1}) - \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1} - A \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that
$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right|}{x_j - x_{j-1}},$$
and it follows that no matter how fine the partition $P$ we have
$$|S(f,P) - A| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(f,P) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.
Best Answer
For each partition $P$, we have$$L(\alpha f,P)=\alpha L(f,P)$$and therefore\begin{align}\underline{\int_a^b}\alpha f&=\sup_PL(\alpha f,P)\\&=\sup_P\alpha L(f,P)\\&=\alpha\sup_PL(f,P)\\&=\alpha\underline{\int_a^b}f.\end{align}