Here's an outline...
Show that $x=y$ mod $m$ if and only if $x=y$ mod $m_i$ for $i=1,\dots,k$.
This will show $x=y$ $\Longleftrightarrow$ $f(x)=f(y)$. The "$\Longrightarrow$" shows $f$ is well defined and "$\Longleftarrow$" shows $f$ is one-to-one.
Showing $f(x+y)=f(x)+f(y)$ should be pretty straightforward as well as $f(xy)=f(x)f(y)$ and $f(1)=(1,\dots,1)$. At this point you'll have established that $f$ is a one-to-one ring homomorphism.
The last step is to show $f$ is onto. For this you'll need to use Chinese remaindering:
Suppose $x_i \in \mathbb{Z}/m_i\mathbb{Z}$ for each $i$. You need to find $x$ such that $x=x_i$ mod $m_i$ for each $i$. The hypothesis that the $m_i$'s are pairwise relatively prime guarantees that there is a solution (this is the Chinese remaindering theorem). Thus $f(x)=(x_1,\dots,x_k)$ and so $f$ is also onto.
Edit: I should have read the question more carefully! To show it's a ring homomorphism you just need to verify the homomorphism properties and establish $f$ is well defined.
For showing well defined: Suppose $x=y$ mod $m$. Then $x-y$ is divisible by $m$. You can conclude that $x-y$ is divisible by $m_i$ since each $m_i$ is a divisor of $m$. Thus $x=y$ mod $m_i$.