I'm able to follow the proof:
dim range $T'$
= dim$W'$ - dim null $T'$
= dim $W$ - dim(range $T$)$^0$
= dim range $T$
However, I'm looking for a better explanation on why they are equivalent. I think my problem stems from still having an unsatisfying understanding of how exactly $T$ and $T'$ are related or affect each other. A better way to phrase my question might be "Do the two ranges happen to have an equal number of dimensions, or is there some underlying property that is causing this?"
Here, I'm using $T'$ to denote the dual map of $T$, and the notation $U^0$ to denote the annihalator of $U$.
The idea behind the adjoint operator is that you can either first go from $V$ to $W$ via $T$ and then apply an element of $W'$ to the result, or you can instead go from $W'$ to $V'$ via $T'$ and then apply that to an element of $V$. Both give the same result.
As астон says in his comments, the underlying reason for the equality of the range dimensions is the connection between the images and kernels of the adjoint operators: $$\begin{align}(\operatorname{im}T)^0&=\ker(T')\\\operatorname{im}(T')&=(\ker T)^0.\end{align}$$ These equalities follow directly from the definition of the adjoint $$(T'\mathbf\beta)[\mathbf v]=\mathbf\beta[T\mathbf v].\tag{1}$$ If we express the above definition of adjoint in matrix form, the connections become fairly obvious. Recall that if we choose dual bases for $V$ and $V'$, we can represent elements of $V'$ as row vectors for which $\mathbf\alpha[\mathbf v]$ becomes the matrix product $$\pmatrix{\alpha_1&\cdots&\alpha_n}\pmatrix{v_1\\\vdots\\v_n}.$$ Applying a transformation to an element of a dual space corresponds to right-multiplying a row vector by a matrix. Using dual bases for $W$ and $W'$ as well, the expression $\mathbf\beta[T\mathbf v]$ then becomes the matrix product $$\pmatrix{\beta_1&\cdots&\beta_m}\pmatrix{t_{11}&\cdots&t_{1n}\\\vdots&\ddots&\vdots\\t_{m1}&\cdots&t_{mn}}\pmatrix{v_1\\\vdots\\v_n}$$ and equation (1) is just a restatement of the associative law for matrix multiplication. In this representation, the matrices of $T$ and $T'$ are identical. We can choose bases for $V$ and $W$ (and their dual bases for $V'$ and $W'$) in which the matrix of $T$ has a particularly simple form*: $$\pmatrix{\beta_1&\cdots&\beta_m}\left(\begin{array}{c|c}I_r&0\\\hline0&0\end{array}\right)\pmatrix{v_1\\\vdots\\v_n}$$ where $r=\operatorname{rank}T$. It should be obvious from this product that $\dim\operatorname{im}T=\dim\operatorname{im}T'=r$. The equalities between kernels and annihilators of the adjoint operators can also be seen by inspection of this product.
* Choose a basis $\mathbf{v}_{r+1},\dots,\mathbf{v}_n$ for $\ker T$ and extend it to a basis $\mathbf{v}_1,\dots,\mathbf{v}_n$ of $V$. Let $\mathbf{w}_i=T\mathbf{v}_i$ for $i=1,\dots,r$ and extend to a basis $\mathbf{w}_1,\dots,\mathbf{w}_m$ of $W$.