On this wikipedia entry https://en.wikipedia.org/wiki/Bilinear_form#Different_spaces it tells us that if $B: V \times W \to K$ is a bilinear map, then
In finite dimensions, [a perfect pairing] is equivalent to the pairing being nondegenerate (the spaces necessarily having the same dimensions).
My question is why does non-degeneracy imply that the induced linear map from $V$ to $W^*$ is an isomoprhism? And, why do $V$ and $W$ necessarily have the same dimension?
You need non-degeneracy in both arguments.
For example, let $\phi\in V^\ast$ be non-zero and define $B\colon K\times V\to K, \,B(\lambda,v)=\phi(\lambda v)$. Then $B(\lambda,v)=0$ for all $v\in V$ implies $\lambda=0$, but $B$ is not a perfect pairing if $\dim(V)\geq 2$.
Now assume that $B$ is non-degenarate in both arguments. Denote by $A\colon V\to W^\ast$ the operator given by $Av(w)=B(v,w)$ for $v\in V$, $w\in W$. This operator is injective since $B$ is non-degenerate in the first argument: $Av=0$ iff $Av(w)=0$ for all $w\in W$ iff $B(v,w)=0$. In particluar, $\dim V\leq \dim W^\ast=\dim W$.
Using non-degeneracy in the second argument, one obtains the converse inequality $\dim W\leq \dim V^\ast=\dim V$. Thus, $\dim V=\dim W$.
Now basic linear algebra yields that $A$ is an isomorphism.