Here's a visual solution. Suppose these are your three generators in $\pi_1(\mathbb C-\{-1,0,1\};2)$:
I will denote them as (inner to outer) $a,b,c$. The analogous generators in $\pi_1(\mathbb C - \{0,1\};4)$ will be $\alpha,\beta$.
Here is the generator $a$ and its image under $f$: 
This one is easy to see as being $\alpha$.
Here is the generator $b$ and its image under $f$: 
Notice that it wraps around $0$ twice, but $1$ only once. Using this and some deformations of the curve, one finds that this is $\beta\alpha^{-1}\beta$.
Finally, here is $c$ and its image under $f$: 
This wraps around $0$ and $1$ twice, and it is in fact $\beta^2$.
To summarize, we have $\pi_1(\mathbb C-\{-1,0,1\};2) = \langle a,b,c\mid\rangle, \pi_1(\mathbb C - \{0,1\};4) = \langle \alpha,\beta\mid\rangle$, and $f_*:\langle a,b,c\mid\rangle\to\langle \alpha,\beta\mid\rangle$ given by $f_*(a)=\alpha,f_*(b)=\beta\alpha^{-1}\beta,f_*(c)=\beta^2$.
It remains to argue why this is injective. You can do this by showing the kernel is trivial. There should be a formal argument for this, but I am really tired and will add that in post-script.
Post-script:
I've been trying to think of a formal argument, but I've come up blank. I'll give some direction towards one though.
We want to show that there is no "word" in terms of $a,b,c$ that evaluates to the identity once we apply $f_*$. (For sake of simplicity, I will just say that $a=\alpha,b=\beta\alpha^{-1}\beta,c=\beta^2$, removing the need for $f_*$.) Essentially, one argues that the $\alpha^{-1}$ inside of $b$ cannot be isolated and cancelled by $a$. We need a $\beta^{-1}$. One way to make a $\beta^{-1}$ is to take $b^{-1}$, but then multiplying $bb^{-1}$ is pointless. Another way is to take $bc^{-1} = \beta\alpha^{-1}\beta^{-1}$. But then $bc^{-1}b = \beta\alpha^{-2}\beta$, leaving the $\alpha$ term "trapped". This may or may not be sufficiently rigorous, but it shouldn't be too difficult from here on.
