Found the answer to my question as well as some new stuff. For starters, whilst $\widetilde{g}=\left(\nabla\vec{v}\right)\left(\nabla\vec{v}\right)^T$ is a correct equation, it isn’t what I’m looking for. The correct usage of this would be to describe the surface of a sphere in 3D with the coordinates u and v. The vector gradient works properly with this, and the resulting metric tensor is 2D and has the curvature of the sphere. You can see what’s actually happening here. We take a surface and merely change the coordinate system. But this cannot affect the surface’s curvature though. By default, we were remapping a flat 4D space to 4D space. So, this method cannot create a curved 4D metric with a 4D vector field. We could create any metric we wanted using a higher dimensional vector field due to the Nash embedding theorem, where the dimension required for an nD metric tensor is at most $\frac{n\left(3n+11\right)}{2}$, so 46D for a 4D metric tensor. Obviously, this is the antithesis of what I wanted.
The formula $\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T$ has also been ruled out. The reasons why are as follows. The formula is linear, so
$$\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T=\nabla\left({\vec{v}}_1+{\vec{v}}_2\right)+\left(\nabla\left({\vec{v}}_1+{\vec{v}}_2\right)\right)^T=\nabla{\vec{v}}_1+\nabla{\vec{v}}_2+\left(\nabla{\vec{v}}_1\right)^T+\left(\nabla{\vec{v}}_2\right)^T=\left(\nabla{\vec{v}}_1+\left({\vec{v}}_1\right)^T\right)+\left(\nabla{\vec{v}}_2+\left(\nabla{\vec{v}}_2\right)^T\right)={\widetilde{g}}_1+{\widetilde{g}}_2$$
All metric tensor fields should be possible, and their configurations determine the form of the stress energy tensor. As such, taking this and the linearity, we add various metric tensor fields together to get a metric tensor with only 1 non-zero unique term. We could create 10 of these fields and add them together to create any 4D metric tensor field. Now we need to check if this formula has the capacity to create such a field. We will start, and end, with the following case.
$$g_{11}=f\left(\vec{x}\right)=\sum_{m_1,m_2,m_3,m_4=0}^{\infty}{c_{m_1,m_2,m_3,m_4}x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}}$$
$$v_1=\frac{1}{2}\int{f\left(\vec{x}\right)dx_1}$$
$$\left(\frac{\partial v_1}{\partial x_2}\right)=\frac{1}{2}\left(\frac{\partial}{\partial x_2}\int{f\left(\vec{x}\right)dx_1}\right)=\frac{1}{2}\int{\left(\frac{\partial f\left(\vec{x}\right)}{\partial x_2}\right)dx_1}=-\left(\frac{\partial v_2}{\partial x_1}\right)$$
$$v_2=-\frac{1}{2}\iint{\left(\frac{\partial f\left(\vec{x}\right)}{\partial x_2}\right){dx_1}^2}$$
$$0=\left(\frac{\partial v_2}{\partial x_2}\right)=-\frac{1}{2}\iint{\left(\frac{\partial^2f\left(\vec{x}\right)}{\partial x_2^2}\right){dx_1}^2}=-\frac{1}{2}\sum_{m_1,m_2,m_3,n_4=0}^{\infty}{\frac{m_2\left(m_2-1\right)}{\left(m_1+1\right)\left(m_1+2\right)}c_{m_1,m_2,m_3,m_4}x_1^{m_1+2}x_2^{m_2-2}x_3^{m_3}x_4^{m_4}}$$
Obviously, this formula only equals 0 if no values of $m_2$ other than 0 or 1 are permitted. The same logically applies for the other coordinates. As such, $f\left(\vec{x}\right)$ must have the form
$$f\left(\vec{x}\right)=f_{000}\left(x_1\right)+x_2\left(f_{100}\left(x_1\right)+x_3\left(f_{110}\left(x_1\right)+x_4f_{111}\left(x_1\right)\right)+x_4f_{101}\left(x_1\right)\right)+x_3\left(f_{010}\left(x_1\right)+x_4f_{011}\left(x_1\right)\right)+x_4f_{001}\left(x_1\right)$$
which cannot describe all metric tensor fields. Therefore, $\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T$ doesn’t describe all metric tensor fields. These finding are relevant to all linear formulas. If they cannot produce a field for one unique component, they aren’t general. The search continues! (or ends here)