When we think of the real square root we think of it as multivalued. The real(and complex) square root of $4$ is $\pm 2$.
But when defining the branches of the square root in $\mathbb{C}$ we usually remove the negative axis. This look couterproductive since the most basic extension that we want i.e the negative reals vanishes which are the once we think of when defining $i$.
Are these branches linked somehow or are they seperate things?
On
The real square root of a nonnegative real number $a$ is usually defined as the nonnegative number $b$, which square is $a$. In your example it's $2$ and not $-2$. But the solutions of the equation $x^2-4=0$ are $\pm 2$. Do you see the difference between square root and solutions of an equation? It's not multivalued, but in $\mathbb{C}$ it is. Make a copy of $\mathbb{C}$ and remove, for example, the nonpositive real axis and link them together. You can see a picture of the square root branches here. They are linked together.
If we want to deal with the square root as a function, we must make a choice concerning which square root we have in mind. If we are dealing with non-negative real numbers, then there is a natural choice: we take the non-negative square root.
If we are dealing with complex numbers, then there is no natural choice; it depends upon the context. You wrote that “we usually remove the negative axis”. Indeed, but there are infinitely many choices. Suppose, for intance, that we remove the numbers of the form $ai$, with $a\leqslant0$. Then every remaining complex number $z$ can be written as $r\bigl(\cos(\theta)+i\sin(\theta)\bigr)$, with $r>0$ and $\theta\in\left(-\frac\pi2,\frac{3\pi}2\right)$, and we can define $s(z)=\sqrt r\left(\cos\left(\frac\theta2\right)+i\sin\left(\frac\theta2\right)\right)$. This function $s$ is also a square root, and it is a continuous (and even an analytic) one.
In order to work with a single square root function, we must leave the complex complex plane and work in a Riemann surface instead.