How many $4$-letter strings either start with c or end with two vowels? If the alphabet consists of $\{a,b,c,d,e,f\}$, how many four letter strings either start with c or end with two vowels?
My reasoning was as follows:
Starts with c but does not have vowel in third position + starts with c but does not have a vowel in fourth position + does not start with c but has vowels in the last two positions $$= 1 \cdot 6 \cdot 4 \cdot 6 + 1 \cdot 6 \cdot 6 \cdot 4 + 5 \cdot 6 \cdot 2 \cdot 2 = 408$$
Is my reasoning correct?

Easier to do inclusion-exclusion.
Starts with C + ends with two vowels - (starts with C and ends with two vowels)
$1*6*6*6 + 6*6*2*2 - 1*6*2*2 = 336$
You were okay but a double adding snuck in.  You counted those that start with C and do not have vowels in either the 3rd or 4th position twice.
Because those things have a way of sneaking in and as if this got even the tiniest bit more complicated you'd have many many things to keep track of, I prefer the inclusion-exclusion method.
