There are two things which I wanted to convey. Summarising below (most of the things here are copied from the above comments):

Assuming $f$ and $g$ are positive functions,

$1) $ If we write: $f(n) + o(f(n)) = \Theta (f(n))$

It means for “any” function $g(n) \in o(f(n)),$ there is “some” function $h(n) \in \Theta(f(n))$ such that

$f(n) + g(n) = h(n)$ for all $n.$

(Remember this equation should be true for all $n,$ not only for large $n.$)

(Now, based on this, we should have to write things for first statement.)

$2) $ If we write $o(f(n)) < f(n),$ it means for any function $g(n) \in o(f(n))$ such that $g(n) < f(n)$ for all $n.$

(As in case of equation, this inequality should also be true for all $n.$)

If we write $g(n) = o(f(n))$ then for some $n,$ you can have $g(n) > f(n).$ and since, we should have $g(n) < f(n)$ So, this statement might not be correct for all positive functions $f(n)$ because inequality may not hold for all $n.$

For example, $f(n) = n^2$ and $g(n) = 10n.$ Here we have $g(n) = o(f(n))$ but $g(2) > f(2)$

So, $10n \in o(n^2)$ and $10n < n^2$ should hold for all $n$ but for $n=2$ it does not hold.

We might loosely say $o(f) < f.$

section “Asymptotic notation in equations and inequalities” in Chapter 3

section 2.5