This site has got 2 example for binary addition. The first,
$0110 + 0111$
I can follow. The second,
$1101 + 0101$
is being used as an example where overflow will happen, because the number of columns in the result is more than the number of columns in either of the addends. But, following on from the first example, would $01101 + 00101$ also be valid?
When we refer to overflows in binary arithmetic (note that this can apply to operations besides addition), we are strictly referring to operations with signed (2's complement) operands, where the result has an inconsistent sign. The "overflow" example that you gave is not technically an overflow - it's a carry-out. Why is it not an overflow? Because note that in the example (in 4 bits), the operands are considered unsigned. We have $13 + 5 = 18$ is $1101 + 0101 = (1) 0010$. This is consistent; we have a carry-out, but no overflow.
However, consider the operation $7 + 5 = 12$ in signed 4-bit arithmetic. This translates to $0111 + 0101 = 1100.$ Note that because these are signed values, the result is $-4$. Somehow we added two positive numbers and ended up with a negative number - the carry from the $N - 1$ bit overflowed into the sign bit.
Overflow can also occur when we add two negative numbers and get a positive result. In some cases, we can have both overflow and carry-out. Consider the operation $(-6) + (-8) = -14$. This translates to $1010 + 1000 = (1)0010$. This would translate to a result of $2$, which is positive. Since we got a positive number by adding two negative numbers, this is an overflow! And since the operation also resulted in 5 bits, there was also a carry-out.
The best way to think about this is that overflow happens when (1) we are considering 2's complement signed operands, and (2) there is a carry into the most-significant (signed) bit. Carry-out happens when there is a carry out of the most-significant bit.
In your proposed case, $01101 + 00101 = 10010$, there is no carry-out. If the operands are considered signed, then there is overflow, since $10010_2 = -14$, which is negative. If the operands are not considered signed, then there is no overflow.