Why is this logical formula invalid for expressing the statement "anything bought it not human"? In my first order logic notes, one example that comes up is to express that "anything bought is not human". Is says this can be expressed as:
$\forall x(\exists y \, bought(y,x) \rightarrow \neg human(x))$
where $bought(y,x)$ means y bought x. It explicitly says not to express it as:
$\forall x \exists y( \, bought(y,x) \rightarrow \neg human(x))$
I read the second one as for all x, there exists y, such that if y bought x, then x is not human. I read the first one as for all x, if there exists a y such that y bought x, then x is not human.
So how does the second formula not express the fact that anything bought is not human? They both seem equivalent in this regard.

Since we don't care about the real world meaning, but the underpinning logic, we'll simply use the relation $B$ and univariate $C$ to shorten the math statements.
The first statement is of the form: $\forall x~((\exists y~B(x,y))\to\neg C(x))$.   Anything will not satisfy $C$ if it is $B$-related to something.  
The second statement is of the form: $\forall x~\exists y~(B(x,y))\to\neg C(x))$.    Everything has something such that, the former does not satisfy $C$ if it is $B$-related to the second.

In the universe $\{a,b,c\}$ where $B(a,c), C(a), C(c)$ are the only facts for $B,C$, then $\forall x~\exists y~(B(x,y)\to\neg C(x))$ is true but $\forall x~((\exists y~B(x,y))\to\neg C(x))$ is false. 

Using equivalences:
$${\quad\forall x~((\exists y~B(x,y))\to C(x))\\ \equiv \forall x~((\neg \exists y~B(x,y))\vee C(x))\qquad\text{implication equivalence}\\\equiv \forall x~((\forall y~\neg B(x,y))\vee C(x))\qquad\text{quantifier duality}\\\equiv \forall x~\forall y~(\neg B(x,y)\vee C(x))\qquad~~~\text{quantifier distribution}\\\equiv \forall x~\forall y~(B(x,y)\to C(x))\qquad\quad\text{implication equivalence}}$$
Which is certainly not equivalent with $\forall x~\exists y~(B(x,y)\to C(x))$, the form of the second statement.
