If $A$ is a $2\times2$ matrix, what is $\det(4A)$ in terms of $\det(A)$?

Let's consider your equation $$4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = \det(A) = 16ad - 16bc$$

The first three objects in this are $2\times 2$ matrices while the last 2 are numbers. So clearly these can't all be equal. What you really want to say here is

$$\color{red}{\det(}4A\color{red}{)} = \color{red}{\det\left(\color{black}{4\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right)} = \color{red}{\begin{vmatrix}\color{black}{4a}&\color{black}{4b}\\\color{black}{4c}&\color{black}{4d}\end{vmatrix}} = 16ad - 16bc$$

Then just complete the logic with a final $$=16\det(A)$$ and you're done.

More generally, if $A$ is $n \times n$, $\det(cA) = c^n\det(A)$ for any scalar $c$. This is because the determinant is a multilinear function of its columns.

In fact, one can define the determinant as the unique function from $n \times n$ matrices to scalars that is $n$-linear alternating in the columns, and takes the value $1$ for the identity matrix.

In particular, $\det(4A) = 4^2 \det(A)$.

Another way to think about it is to remember that the determinant of an $n \times n$ matrix gives the (signed) volume of the $n$-dimensional box enclosed by its column vectors. Increasing each side length by a factor of $k$ will increase the volume by a factor of $k^n$.