Alternate reduction from 3SAT to 4SAT? It seems that the standard reduction method you see online from 3SAT to 4SAT is that we let $\phi = (a \lor b \lor c)$ be a 3SAT clause, and so there is an assignment that satisfies $\phi$ iff $\phi' = (a \lor b \lor c \lor z) \land (a \lor b \lor c \lor \neg z)$ is also satisfiable.

Is it just as valid to say that $(a \lor b \lor c)$ is satisfiable iff $(a \lor b \lor c \lor c)$ is satisfiable? This seems like a much simple solution..
It depends on the exact definition of $k$-SAT that's being used.  If $k$-SAT is defined to have clauses of length at most $k$, your reduction is perfect; if it's defined to have clauses with exactly four distinct literals, then you're out of luck.