Say I have a function $$f:\mathbb{R}^+ \rightarrow \mathbb{R}$$ defined as $$f(x)=\sqrt{x}$$
Now, going by the standard procedure by taking some arbitrary element $\mathbf{y}\in{R}$ and solving to get some $\mathbf{x=g(y)}$, we end up with $x=y^2$. And then we see that for all $$\mathbf{y\in{R}},$$ $$\mathbf{x\in{R^+}}.$$ Thereby it seems that the function is $surjective.$ I understand that it is trivial to understand from definition that the function is not $onto$. So does that mean I have to check the initial conditions everytime I see a $surd$ in a function before proceeding as above, or am I missing something?
Surds don't have anything to do with the issue. Let's take a look at what you did: you fixed $y \in \mathbb{R}$ and tried to find $x \in \mathbb{R}^+$ such that $y = \sqrt{x}$. Then you squared both sides and obtained $y^2 = x$.
But since squaring is not injective, we have that $y^2 = x$ is a necessary condition for $y = \sqrt{x}$, but it is not sufficient. That's why right now you only know that any number $x$ that works (i.e. such that $y = \sqrt{x}$) must be equal to $y^2$, but you don't know if $y^2$ will work itself. To complete the argument, you have to check it manually by plugging in: is it true that $y = \sqrt{y^2}$? If you do, you'll see that it's not always true, namely - it fails for negative $y$.
So in general, when you solve an equation, such as $f(x) = y$, and conclude from the equation that $x \in X$ for some set $X$ - i.e. if you prove that for all $x, y$ we have that
$$y = f(x) \implies x \in X$$
- then you still have to check which elements of $X$ actually satisfy the equation.
You don't have to do that if all the transformations you make are equivalences, i.e. if you prove that for all $x, y$ we have that
$$y = f(x) \iff x \in X,$$
then you can immediately tell that $X$ is the solution set. Of course in your specific case $X = \{ y^2 \}$.