The shorthand notation is tripping you up. Let's say that in a particular coordinate chart $x$, your vector components take the form $V^\mu (x)$. You now decide to differentiate these components, to obtain an expression $$\frac{\partial}{\partial x^\nu}\big[ V^\mu(x)\big]$$
Now you wish to change to a different coordinate chart $y$. The components of $V$ transform as $$V^\mu(x) \mapsto J^\mu_{\ \ \alpha}(y) \hat V^\alpha(y)$$ where $J^\mu_{\ \ \alpha}(y)\equiv \frac{\partial x^\mu (y)}{\partial y^\alpha}$ is the Jacobian of the transformation. Plugging that in yields
$$\frac{\partial}{\partial x^\nu}\left[J^\mu_{\alpha}(y)\hat V^\alpha(y)\right]$$
The problem now is that the quantity inside the brackets is a function of the coordinates $y$, so you cannot differentiate them with respect to the coordinates $x$. Instead, you need to transform your derivative operator: $$\frac{\partial}{\partial x^\nu} \mapsto \big(J^{-1}(y)\big)^\beta_{\ \ \nu} \frac{\partial}{\partial y^\beta}$$
Now you can expand everything out using the product rule. To recover the shorthand notation, you can drop the arguments $(x)$ and $(y)$ from all of the functions, write the Jacobian as $\partial x/\partial y$, and replace $y$ with $x'$. From there, replace $\frac{\partial}{\partial x^\mu}$ with $\partial_\mu$ and $\frac{\partial}{\partial x'^\mu}$ with $\partial_{\mu'}$. (Can you see how this gets confusing for beginners?)