Simple example: Suppose you have a hash that is 136 bits but the last 8 bits are always 0.
Clearly it is not more secure than just the first 128 bits.
In practice it will be less obvious but it means that the 136 bit long output will be one of only $2^{128}$ different values.
The extra bits are required to guarantee 128 bit security and cannot be omitted.
To continue the analogy, assume that the 8 extra zero bits are mixed with the rest of the 128 bits. You simply don't know which bits to omit.
To give a possibly more visual, short example consider the following fictitious hash outputs:
$100, 110, 011, 010$
Since there are only 4 possibilities the hash could not have a security level greater than $2^2$. (You could simply attack it by guessing one out of the $2^2$ values.)
However, the output is 3 bits long and you cannot remove any of the bits. If you did you would have less than $2^2$ possibilities left.