Suppose that $g(n)$ is $o(f(n))$, and let $h(n)=f(n)+g(n)$; you want to prove that $h(n)$ is $\Theta(f(n))$.
To show that $f(n)$ is $\Theta(h(n))$, you must show that there are positive constants $C_1,C_2$, and $n_0$ such that $C_1|f(n)|\le|h(n)|\le C_2|f(n)|$ for all $n\ge n_0$.
Note: My $n_0$ corresponds to your $n'$. Note that in your post you have the inequalities involving $n$ and $n'$ backwards: you want something to be true for all $n$ from some point on, not just for all $n$ up to some point. Your definitions of $o$ and $\Theta$ are seriously off in other respects as well; you should check them against your text or this article.
Unpacking $h$, we want $C_1|f(n)|\le|f(x)+g(x)|\le C_2|f(n)|$ for all $n\ge n_0$. Since $g(n)$ is $o(f(n))$, we know that for each $\epsilon>0$ there is an $n_\epsilon$ such that $|g(n)|<\epsilon|f(n)|$ whenever $n\ge n_\epsilon$. Take $\epsilon=\frac12$, and let $n_0=n_{1/2}$. Then whenever $n\ge n_0$ we have $|g(n)|\le\frac12|f(n)|$, so
$\frac12|f(n)|=|f(n)|-|g(n)|\le|f(x)+g(x)|\le|f(n)|+|g(n)|\le\frac32|f(n)|\;,$
and we see that $C_1=\frac12$ and $C_2=\frac32$ do the trick with this value of $n_0$.