To see that your answer $4^k-[4+6(2^k-2)+4(3^k-10)]$ is wrong, just try plugging in some small values of $k;$ for $k=1,2,3,4$ you should get $0,0,0,24$ if your formula is correct.
The in-and-out formula is designed for problems like this. For four sets $A_1,A_2,A_3,A_4\subseteq E$ the formula is
$$|E\setminus(A_1\cup A_2\cup A_3\cup A_4)|=$$
$$|E|$$
$$-(|A_1|+|A_2|+|A_3|+|A_4|)$$
$$+(|A_1\cap A_2|+|A_1\cap A_3|+|A_1\cap A_4|+|A_2\cap A_3|+|A_2\cap A_4|+|A_3\cap A_4|)$$
$$-(|A_1\cap A_2\cap A_3|+|A_1\cap A_2\cap A_4|+|A_1\cap A_3\cap A_4|+|A_2\cap A_3\cap A_4|)$$
$$+|A_1\cap A_2\cap A_3\cap A_4|.$$
If $E$ is the set of all functions from $[k]=\{1,\dots,k\}$ to $[4]=\{1,2,3,4\},$ and $A_i$ is the set of all functions $f\in E$ such that $i$ is not in the range of $f,$ then
$$|E|=4^k,$$
$$|A_1|=|A_2|=|A_3|=|A_4|=3^k,$$
$$|A_1\cap A_2|=|A_1\cap A_3|=|A_1\cap A_4|=|A_2\cap A_3|=|A_2\cap A_4|=|A_3\cap A_4|=2^k,$$
$$|A_1\cap A_2\cap A_3|=|A_1\cap A_2\cap A_4|=|A_1\cap A_3\cap A_4|=|A_2\cap A_3\cap A_4|=1^k=1,$$
$$|A_1\cap A_2\cap A_3\cap A_4|=0^k=\begin{cases}
0\text{ if }k\gt0,\\
1\text{ if }k=0.
\end{cases}$$
Thus, For $k\ge1,$ the number of "onto" functions from $[k]$ to $[4]$ is
$$|E\setminus(A_1\cup a|2\cup A_3\cup A_4)|=4^k-\binom413^k+\binom422^k-\binom431^k+\binom440^k=\boxed{4^k-4\cdot3^k+6\cdot2^k-4}\ .$$
Likewise, the number of "onto" functions from $[k]$ to $[3]$ is
$$3^k-\binom312^k+\binom321^k+\binom330^k=3^k=\boxed{3^k-3\cdot2^k+3}\ne3^k-10.$$
