Suppose we have that $\bf{F}=\vec{\nabla}$ ${f}$ and $\bf{G}=$ $\vec{\nabla}g$ are three gradient vector fields in $\mathbb{R}^n$, such that $\bf{H}=$ $c_1\bf{F}+$ $c_2\bf{G}$ for some $c_1, c_2 \in \mathbb{R}$. To prove that $\bf{H}$ is also a gradient vector field, we can probably proceed as follows:
$\bf{H}=$ $c_1\vec{\nabla}f + c_2\vec{\nabla}g = c_1 (\frac{∂f}{∂x_1}, ..., \frac{∂f}{∂x_n})+c_2 (\frac{∂g}{∂y_1}, ..., \frac{∂g}{∂y_n})=(c_1 \frac{∂f}{∂x_1} + c_2 \frac{∂g}{∂y_1},..., c_1 \frac{∂f}{∂x_n} + c_2 \frac{∂g}{∂y_n})$. Now we can conclude that since $\bf{H}$ is a linear combination of functions $\bf{F}$ and $\bf{G}$, the linear combinations of the partial derivatives are also corresponding partial derivatives of $\bf{H}$, and hence $\bf{H}$ $=\vec{\nabla}h$ for some function $h$.
While writing this proof, I realized that perhaps, WLOG, all $y_i$'s in the derivatives of $g$ should be changed to $x_i$'s.
Do you think this proof is correct?
Not just perhaps or WLOG but definitely the coordinates should all be the same, $x_i$.
What you've written is not a proof that $\mathbf H$ is a gradient vector field; to do that, you need to show $\mathbf H=\vec\nabla h$ for some function $h$. As kobe has pointed out in a comment, the appropriate function is $h=c_1f+c_2g$, so you need one more step where you rewrite $c_1\frac{\partial f}{\partial x_i}+c_2\frac{\partial g}{\partial x_i}$ as $\frac{\partial}{\partial x_i}\left(c_1f+c_2g\right)$.