Basis vectors and Basis Covectors 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?
 A: 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 $\frac{\partial}{\partial x^j}\bigg|_p$
