Why is the inverse of a sum of matrices not the sum of their inverses? Suppose $A + B$ is invertible, then is it true that $(A + B)^{-1} = A^{-1} + B^{-1}$?

I know the answer is no, but don't get why.
Let us look at the often-forgotten $1 \times 1$ matrices over $\mathbb{R}$, which is another name for the real numbers themselves. Then your statement translates to the statement that if $x,y$ are real numbers that aren't zero, then
$$ \frac{1}{x + y} = \frac{1}{x} + \frac{1}{y},$$
which is clearly wrong.
It's just that the operation of taking inverses isn't well-behaved when it comes to sums.

This isn't surprising, because taking inverses is defined in terms of multiplication and NOT in terms of addition. There is no reason really why inverses should act in some favorable way on sums.

For instance, if $A$ is some invertible matrix, then what is the inverse of $A+A$? Is it $A^{-1}+A^{-1}$? No. You have that $A+A=2A$, whose inverse is $1/2 A^{-1}$, and this is not the same as $A^{-1}+A^{-1}=2A^{-1}$.

What's more is that if $A$ and $B$ are invertible matrices, then their sum $A+B$ need not even be invertible, as others have pointed out: It is not even always possible to find $(A+B)^{-1}$, even though both $A^{-1}$ and $B^{-1}$ exist.

What does hold, which you probably already know, is that (for invertible $A$ and $B$), we have this nice multiplicative property:

$(AB)^{-1} = B^{-1}A^{-1}.$