The definition you give in the quesition is very imprecise so it's understandable that you've misunderstood it. The formal definitions are given at our reference question on NP.
A decision problem is in NP if there's some polynomial $p$ and an ordinary, deterministic polynomial-time Turing machine $M$ such that:
For every "yes" instance $x$, there is some $y$ with $|y|\leq p(|x|)$ such that $M$ accepts the string $x,y$;
For every "no" instance $x$, $M$ rejects all strings $x,y$.
So, for example, 3-COLOURABILITY is in NP: the string $y$ is just a description of a $3$-colouring and $M$ accepts $x,y$ if $y$ is a valid $3$-colouring of the graph $x$ and rejects otherwise. The point is not that you can "verify 'yes' instances" but, rather, that every "yes" instance has a certificate $y$ that you can use to easily prove that it really is a "yes" instance.
But there's a significant asymmetry in the definition. If you give $M$ a "yes" input and a certificate $y$, then $M$ will say "I agree – that's a 'yes' instance." However, there's no input you can give it that will cause it to say "I agree – that's a 'no' instance." For any certificate you try, it will just say "That doesn't convince me it's a 'yes'." You could, of course, try all possible certificates. That would certainly prove that you had a "no" instance but there's a problem: there are exponentially many possible certificates to try, so trying all of them doesn't give you a polynomial-time algorithm.