For a set of linearly independent vectors $\{u_1, u_2 ... u_n\}$, the Gram-Schmidt process for obtaining orthogonal vectors $\{v_1, v_2 ... v_n\}$ is:
$v_1 = u_1 \\ v_2 = u_2 - proj_{v_1} u_2 \\ ... \\ v_n = u_n - \sum_{0}^{n-1} proj_{v_n}u_n$
Is there any difference in result between this process, and:
$v_1 = u_1 \\ v_2 = u_2 - proj_{u_1} u_2 \\ ... \\ v_n = u_n - \sum_{0}^{n-1} proj_{u_n}u_n$
I view the Gram-Schmidt process as a method of removing alike components from a vector until all vectors are orthogonal, so intuitively I feel like it should result in a valid basis albeit different.
Yes, this is different, and the latter process typically does not give you an orthogonal vectors. For instance, let's take the vectors $u_1=(1,0,0)$, $u_2=(1,1,0)$, and $u_3=(1,0,1)$ in $\mathbb{R}^3$. Gram-Schmidt gives $$v_1=u_1=(1,0,0),$$ $$v_2=u_2-u_1=(0,1,0),$$ and $$v_3=u_3-(u_1+0)=(0,0,1).$$
On the other hand, your proposed process gives the same values for $v_1$ and $v_2$, but instead gives $$v_3=u_3-\left(u_1+\frac{1}{\sqrt{2}}u_2\right)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},1\right).$$ This vector $v_3$ is not orthogonal to $v_1$ or $v_2$.
What's going on here is that to define $v_3$, you really want to take $u_3$ and subtract its projection onto the subspace spanned by $u_1$ and $u_2$ (or equivalently, the subspace spanned by $v_1$ and $v_2$). If you want to describe this as a sum of projections onto 1-dimensional subspaces, you need those subspaces to be orthogonal. That is, you need to project onto $v_1$ and $v_2$ which you have arranged to be orthogonal, rather than $u_1$ and $u_2$ which may not be orthogonal.