You're half-correct. By stars and bars, the number of solutions to $$x_1+x_2+x_3+x_4=8$$ is exactly $\binom{8+4-1}{4-1}=165$. I don't understand however why numbers like $8=0008$ don't count, since, as Thomas Andrews pointed out in the comments, the sum of its digits add up to $8$.
However, for $16$, our approach cannot be exactly the same, since a solution to $x_1+x_2+x_3+x_4=16$ could be $(x_1,x_2,x_3,x_4)=(1,2,3,10)$, but $10$ cannot be a digit, as those are always between $0$ and $9$. Since only one can be $10$ or larger, we can simply count the number of solutions where one is $10,11,\cdots,16$. If one is $10$, then we count the number of solutions to $x_1+x_2+x_3=16-10=6$. Use stars and bars again and we get $\binom{6+3-1}{3-1}$. Now we multiply by $4$ since we now chose $x_4=10$ but we also need to count $x_1,x_2,x_3=10$. So we get $4\binom{6+3-1}{3-1}$. Using this principle we get $$4\left(\binom{6+3-1}{3-1}+\binom{5+3-1}{3-1}+\cdots+\binom{0+3-1}{3-1}\right)=4\cdot84=336$$ The total amount of solutions to $x_1+x_2+x_3+x_4=16$, including those we don't want, are (again, stars and bars) $\binom{16+4-1}{4-1}=969$. Now substract the amount of solutions we don't want, and we get $969-336=633$ possibilities.
As Eric noted in a comment, you're not taking into account the fact that there are four different positions for the fourth digit. The correct calculation is
$$
10^4-4\cdot10\cdot10+3\cdot10=9630\;,
$$
where the last term corrects for the fact that the second term counts each string of four identical digits $4$ times. Alternatively, not counting them in the second term, you could also write
$$
10^4-4\cdot10\cdot9-10=9630\;.
$$
Best Answer
This problem can be reduced to solving
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9, \quad x_1 \neq 0, \quad x_1 \neq 9$$
First, we solve
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 15$$
This is a standard stars and bars problem, with solution $\binom{15+3-1}{15} = 136$. Now which solutions are wrong? Well, first of all those with one of them above $10$. But if, say, $x_1 \geq 10$, then we can write $x_1' = x_1 - 10$ and $x_2' = x_2, x_3' = x_3$ to reduce the problem to
$$x_1' + x_2' + x_3' = 5, \quad 0 \leq x_i' \leq 5$$
This is again a standard stars and bars problem, with solution $\binom{5+3-1}{5} = 21$. Since exactly one of $x_1,x_2,x_3$ could be above $10$, this gives $3 \cdot 21 = 63$ wrong solutions in total. So there are $136 - 63 = 73$ solutions to
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9$$
Finally, we need to subtract $4$ solutions with $x_1 = 0$ (i.e. $(x_2,x_3) \in {(6,9),(7,8),(8,7),(9,6)}$), but also subtract $7$ solutions for $x_1 = 9$ (i.e. $(x_2,x_3) \in {(0,6),\ldots,(6,0)}$). This gives $62$ solutions in total.