I'm interpreting your question as "In what sense is the equivariant Euler class of a torus representation, thought of as a vector bundle on the point, just the product of the weights of the representation?" Because Euler classes multiply under direct sum, it's enough to answer this question for a 1-d representation.

Now, it's a general principle that if a space is nice in some sense, there will be a bijections between complex line bundles on that space and $H^2(X;\mathbb{Z})$ via first Chern class.  One way to think of this correspondence is that a line bundle is defined by an element of $H^1(X;\mathcal{O}^*)$ where $\mathcal{O}^*$ is the nonvanishing elements of whatever sheaf of functions is relevant.  As long as the sheaf cohomology of $\mathcal{O}$ is boring, the boundary map in the long exact sequence for the exponential sequence $\mathbb{Z}\to \mathcal{O}\to\mathcal{O}^*$ induces this isomorphism.

So, now what does this mean equivariantly?  Well, remember that equivariant cohomology is the cohomology of a space.  In the case of a point, it is the cohomology of the classifying space $BT$.  

On the other hand, a line bundle on $BT$ is the same thing as a 1-dimensional representation of $T$; you use the standard associated bundle construction.  

So, by the argument above, we get an identification between characters of $T$ and $H^2_T(pt;\mathbb{Z})$.  This is being used implicitly when you make a statement like "the equivariant Euler class of the normal bundle at the fixed point (i.e. the tangent space at that point) is (upto a sign) the product of the weights of the action of the lie algebra of the Torus on the tangent space at that point," but having fixed this isomorphism, the statement above becomes tautological.