You are correct that $$(P \rightarrow Q) \land (Q \rightarrow R) \not\equiv P\rightarrow R$$
It is true that $$\Big((P\rightarrow Q) \land (Q\rightarrow R)\Big) \rightarrow (P \rightarrow R)$$ (You can confirm that this is a tautology by consulting its truth-table.)
But if $Q$ is false, and P is true, and R is true, the left hand side is false, but the righthand side is true, so the implication is not bidirectional.
The argument :
$t \rightarrow a, a \land m $, therefore : $m \land t$
is not valid, because is it a little bit "more complicated" form of the fallacy known as Affirming the consequent :
(if P, then Q) and Q, therefore P.
To understand it, we have to re-phrase a little bit the three statements :
i) : "An interesting teacher keeps me awake" must be rewritten as :
"If My teacher is interesting, then I stay awake"
that is, in symbols : $t \rightarrow a$.
ii) : "I stay awake in Discrete Mathematics class" as :
"I am in Discrete Mathematics class and I stay awake"
that is, in symbols : $m \land a$.
iii) : "My Discrete Mathematics teacher is interesting", i.e. :
"I am in Discrete Mathematics class and My teacher is interesting"
that is : $m \land t$.
Best Answer
You need to look at $p$, not $\lnot p$. In your case, if $q$ is false and $\lnot p$ is false (i.e. $p$ is true), $q$ unless $\lnot p$ is false, so in order of ($p$, $q$, statement) it is (true, false, false) which is the same as $p \rightarrow q$
They are different in English but the same in math. If the politician is not elected ($p = F$), and he lowered tax ($q = T$), he will lower tax if he is elected is true while he will lower tax unless he is not elected sounds like a false. The confusion comes because in mathematics, given the conditions above, (he will lower tax unless he is elected) is true.
I think a better way to explain $q$ unless $\lnot p$ is "$q$ unless $\lnot p$, but if $\lnot p$ really happens then $q$ doesn't matter". (like what @skyking said)
Actually I believe Are these two statements equivalent? is the same question.