Compound propositions as assertions? According to comments on my previous question, compound propositions are not assertions; i.e. the statement "$p \vee q$" does not mean "$p$ (is true) or $q$ (is true)", and it does not mean "$(p$ or $q)$ is true".From my understanding, "$p \vee q$" is just an un-evaluated expression, just like $2+3$.
However, my book says

The statement $p \implies q$ is called a conditional statement because $p \implies q$ asserts that $q$ is true on the condition that $p$ holds

which seems to say that we can interpret "$p \implies q$" as "$p$ (is true) $\implies q$ (is true)". This appears to directly contradict the comments in the other question, especially since we can write a conditional statement as a disjunction. Indeed, if we assume that "$p \vee q$" means "$p$ (is true) or $q$ (is true)", the truth tables match. If we assume  "$p \vee q$" means "$p$ (is false) or $q$ (is false)" instead, the truth table does not match with $p \vee q$; "$p$ (is false) or $q$ (is true)" (or the other way around) doesn't match either. 
So my question is:
Is there a contradiction between the comments on my previous question and the book?
If anyone is wondering, my book is "Discrete Mathematics and Its Applications" $6^{th}$ edition by Kenneth H. Rosen.

With regards to the bulk of your question before your example:
The conditional statement asserts asserts that $q$ is true on the condition that $q$ holds, but the statement itself can be wrong. I can say that my statement $r$ is "if I never study ($p$), then I will always get 100% on my exams ($q$)". Then, when I refuse to study and fail the exam, $r$ itself is shown to be false.
In other words, $P \implies Q$ tries to assert something, but it is an unevaluated expression as much as $P \vee Q$ is.
I also used the Rosen textbook (7th edition, doesn't matter)
