If I define a vector so that:
$$ \vec v = v^i e_i = v_i e^i $$
Where $ e_i e^j = \delta _i^j$, then its fair to say that $e_i$ and $e^i$ are dual vectors to eachother. Now from this, we can show that the metric tensor $g_{ij} $ can convert between the components of the original basis and the new basis. So we can see this by first directly computing the vector product:
$$ \vec v \cdot \vec v = v^iv^j e_i \cdot e_j = v^i \color{red}{v^j g_{ij}}$$
Then doing it again using the dual representation:
$$ \vec v \cdot \vec v = v^iv_j e_i \cdot e^j = v^i v_j \delta_i^j = v^i\color{red}{v_i} $$
So by equating the red terms here, we can see immediately that:
$$ v_i = v^j g_{ij} $$
Now if the metric tensor transforms the dual component in this way, what I'm trying to figure out is, how can I express my basis vector in terms of the dual basis vector.
My initial inclination is to expect that the opposite thing should happen, my reasoning there is:
$$ \vec v = v_ie^i = v^j \color{blue}{g_{ij}e^i} = v^j\color{blue}{e_j} $$
So then immediately, we can say that:
$$ e_j = g_{ij}e^i $$
Now this is slightly problematic for me though, because it would imply that differential forms are related to basis vectors such that $ {\partial \over \partial c^i} = g_{ij} dc^i $ which I haven't seen anywhere. so I suppose I'm asking if there are any errors in my calculations. The only one I can imagine is that differential forms are covectors, but if the metric tensor converts vectors into covectors, then why can't it do the same to the basis vectors as well?
The problem in what you wrote is that it is not formally correct. Following Jeffrey M. Lee - "Manifolds and differential geometry":
Take $(V,g)$ to be a scalar product space, now since $g$ is nondegenerate, there exists a linear ismorphism $g_\flat:V\to V^*$ defined by: $$w\mapsto[v\mapsto g_\flat(v)(w)]$$ with $$g_\flat(v)(w)=g(v,w)$$ in other words: $$g_\flat(v)=g(v,\cdot)\in V^*$$ We can even define its inverse by $g^\sharp:V^*\to V$, by forcing it to be an isometry bt defining a scalar product on $V^*$: $$g^*(\alpha,\beta)=g(g^\sharp(\alpha),g^\sharp(\beta))$$ Under this, the dual basis $(e^1,...,e^n)$ corresponding to an orthonormal basis $(e_1,...,e_n)$ will also be orthonormal.
As a matter of definition, we will indicate : $$g_\flat(v)=:v^\flat:=\flat v$$ The maps $g_\flat$ and $g^\sharp$ are called musical isomorphisms.
Take now an arbitrary basis $(f_1,...,f_n)$ of $V$ and let $(f^1,...,f^n)$ be the dual basis for $V^*$. The components of $G$ are $g_{ij}=g(f_i,f_j))$ so if $v=v^if_i$ and $w=w^if_i$ then: $$g(v,w)=g(v^if_i,w^jf_j)=v^iw^jg(f_i,f_j)=v^iw^jg_{ij}$$
Now, since there must be a matrix $(A_{ij})$ such that $\flat f_i=A_{ki}f^k$, and $$g_{ij}=g(f_i,f_j)=(\flat f_i)(f_j)=A_{ki}f_j=A_{ki}\delta^k_j=A_{ij}$$ Then we have $$\flat f_i=g_{ik}f^k$$ Thus, for $v=v^if_i$ we have $$\flat v=v^jg_{ji}f^i$$ in such a way that the components of $\flat v$ are: $$(\flat v)_i=v^jg_{ji}$$ So, the correct relations that you should have in mind is the latter here. I hope this will make some clarification.
EDIT:: In a Riemannian Manifold $M$ you have a metric tensor $\forall p\in M$: $$g_p:T_pM\times T_pM\to\mathbb{R}$$ and then $g_p\in T_pM^*\otimes T_pM^*$ so that it has the form in a local chart $(U,x)$: $$g_p=g_{ij}|_pdx^i|_p\otimes dx^j|_p$$ where $g_{ij}|_p=g_p\left(\frac{\partial}{\partial x^i}\bigg|_p,\frac{\partial}{\partial x^j}\bigg|_p\right)$. If you now consider: $$g_p\left(\frac{\partial}{\partial x^k}\bigg|_p,\cdot\right)=g_{ij}|_pdx^i|_p\left(\frac{\partial}{\partial x^k}\bigg|_p\right)dx^j|_p=g_{ij}|_p\delta^i_kdx^j|_p=g_{kj}|_pdx^k|_p=:\omega_j|_p$$ And nothing garantees that this $\omega_j|_p$ is $\frac{\partial}{\partial x^j}\bigg|_p$