I understand the concept of blowing up of an affine scheme along an arbitrary center but when it comes to compute strict transforms I have some difficulties in very simple examples. I would appreciate if someone can help me. Being specific, look at the affine variety $\mathbb{A}_k^2$ and let's blow up the ideal $I=(y^2+x^3,x^2(1+y))$. The result is a scheme/variety $X$ admitting an open cover by two affine pieces $U_0$ and $U_1$ with coordinate rings $k\Big[x,y,\dfrac{y^2+x^3}{x^2(1+y)}\Big]$ and $k\Big[x,y,\dfrac{x^2(1+y)}{y^2+x^3}\Big]$, respectively.
What happens if we try to compute equations to the strict transform of the curve $C=V(y^2+x^3)$?
Notice that IF $I=(x,y)$ we simply (in one chart) put $y^2+x^3=(x'y')^2+x'^3$ and clean the $(x')^2$ factor.
What is the corresponding procedure in a "general" situation like the previous one?
For your specific example, note that in your case the strict transform of the curve $C$ is just going to be itself, because you are blowing up an effective Cartier divisor on $C$.
I'm not sure of any shortcut - probably your best bet would just be to check your definitions. Namely, suppose you would like to blow up the ideal $I=(f_0,\ldots,f_k)$ in $\operatorname{Spec}(A)$. Recall that the blow up is given by $\operatorname{Proj} \bigoplus_{i\ge 0} I^i$. You have a surjection of graded rings $A[T_0,\ldots,T_k] \rightarrow \bigoplus_{i\ge 0} I^i$ mapping $T_i \mapsto f_i$, which induces a closed embedding $Bl_I A \hookrightarrow A \times \mathbb{P}^k$. Then by figuring out the kernel, you can figure out what your blow up looks like.
Suppose you have computed the blow up of $X$ w.r.t. $I$ and you want to compute the strict transform of a subscheme $Y$ of $X$, then what you really do is just to proceed as before - but being careful with your equations.
Let's do an example (and I'll pick a slightly simpler one than the one you gave just to show how the computations work in general): Let's suppose we are blowing up $(x^2,y^2)$ in $\mathbb{A}^2$ and you want to compute the strict transform of $y^2=x^3$. Like before, you should also expect the strict transform to be isomorphic to $y^2=x^3$, since you are blowing up an effective Cartier divisor on $y^2=x^3$. But let's see how the equations play out.
Let $A=k[x,y]$. You have a surjection of graded rings $A[S,T] \rightarrow \bigoplus_i (x^2,y^2)^i$ given by $S\mapsto x^2, T\mapsto y^2$. The kernel of this map is given by $y^2S - x^2T=0$ in $\mathbb{A}^2 \times \mathbb{P}^1$ (Observe that this gives you the Whitney umbrella on an affine chart).
Now the total transform is then given by the zero locus of $y^2S - x^2T = 0$ and $y^2-x^3=0$. To compute the strict transform, you want to compute the closure of the preimage of $\{y^2-x^3=0\} \backslash (0,0)$, which really amounts to finding equations whose zero locus contains the preimage of $\{y^2-x^3=0\} \backslash (0,0)$ as an open dense subset.
Observe that if $S=0$, then we must have $x=0$ and hence $y=0$. Therefore, we can just work on the affine chart $S\ne 0$, and the total transform is given by $y^2 - x^2t = 0, y^2-x^3=0$. So the total transform is given by the ideal $(y^2-x^2t, y^2-x^3) = (y^2-x^3, x^2(x-t))$. We may assume that $x$ is invertible (because we are considering the preimage of $\{y^2-x^3=0\} \backslash (0,0)$, and so you will find that the strict transform is cut out by the ideal $(y^2-x^3, x-t)$ in $k[x,y,t]$, which is clearly isomorphic to what we started with.