How can one reduce 3-CNF-SAT and k-CNF-SAT to each other?
To reduce $k$-CNF to CNF: every instance of $k$-CNF is an instance of CNF, so the reduction is trivial.

To reduce CNF to $k$-CNF, where $k \ge 3$: convert the CNF formula to a circuit that computes it; then convert the circuit to a 3-CNF formula using the Tseitin transformation.  (If your definition of $k$-CNF requires that there be exactly $k$ literals in every clause, then you can pad: introduce $k$ new variables $y_1,\dots,y_k$; for every clause that has less than $k$ literals, add $\lor y_1 \lor y_2 \lor \dots$ to pad it out to exactly $k$ literals; and then add clauses that force all of $y_1,\dots,y_k$ to be false.)

If this is too terse, see our reference question and a good textbook to learn the fundamentals of reductions.