Check that $F:A/\mathfrak p\longrightarrow B/\mathfrak q$ is well defined. Let $A,B$ commutative rings and $\mathfrak p,\mathfrak q$ ideals of $A$ and $B$ respectively. We define $$F:A/\mathfrak p\longrightarrow B/\mathfrak q$$
by $$F(a+\mathfrak p)=f(a)+\mathfrak q$$
where $f:A\longrightarrow B$ is ring homomorphism and $\mathfrak q$ is the ideal generated by $f(\mathfrak p)$. I want to check that $F$ is well defined. The problem is that it looks too simple since usually to check that such functions are well defined is more complicate. So is it correct ?
Let $a+\mathfrak p=b+\mathfrak p$. Since $\mathfrak p$ is an ideal, we have that $a-b\in \mathfrak p$. Therefore, $$F(a+\mathfrak p)=F(a+b-b+\mathfrak p)=F(b+(a-b)+\mathfrak p)=F(b+\mathfrak p).$$
Is it correct ? In fact, even if $A$ is not a commutative ring it works, no ? Since $(A,+)$ is always a commutative group. Can I then deduce that if $A$ is a ring and $I$ is a both side ideal, then all function $$F:A/I\longrightarrow B$$
is well defined ? 
 A: Your solution is correct in the idea, but you skip over important details by using $F(a+\mathfrak{p})$ before actually checking the map exists, so you skip over an important point. In your proof you use an argument that may look circular. If you use the $f$ version, you have
$$
f(a)+\mathfrak{p}=f(a+b-b)+\mathfrak{p}=f(b)+f(a-b)+\mathfrak{p}=f(b)+\mathfrak{p}
$$
You can improve and simplify it by observing that $a-b\in\mathfrak{p}$ implies $f(a)-f(b)=f(a-b)\in f(\mathfrak{p})\subseteq\mathfrak{q}$ and therefore $f(a)+\mathfrak{q}=f(b)+\mathfrak{q}$.
You can also “abstract” it.
Let $f\colon A\to B$ be a ring homomorphism; if $\mathfrak{p}$ is an ideal of $A$ and $\mathfrak{q}$ is the ideal of $B$ generated by $f(\mathfrak{p})$, we can consider the ring homomorphism
$$
g=\pi\circ f\colon A\to B/\mathfrak{q}
$$
where $\pi\colon B\to B/\mathfrak{q}$ is the canonical projection. If $a\in\mathfrak{p}$, then $f(a)\in\mathfrak{q}$ and so $g(a)=0$. Hence $a\in\ker g$.
The homomorphism theorem now says that there is a unique ring homomorphism $F\colon A/\mathfrak{p}\to B/\mathfrak{q}$ such that $F(a+\mathfrak{p})=g(a)=\pi\circ f(a)=f(a)+\mathfrak{q}$.

About your last question, it is not clear what you mean. What is true is that a ring homomorphism $f\colon A\to B$ induces a ring homomorphism $f_I\colon A/I\to B$, for all ideals $I$ such that $I\subseteq\ker f$. This is already contained in the previous statement, because in this case $f(I)=\{0\}$.
A: This is not correct. You have to show the equality $F(b+(a-b)+\mathfrak p) = F(b+\mathfrak p)$: this is, where you actually use different representatives.
In other words: If we have $a+\mathfrak p = b+\mathfrak p$, then $a$ and $b$ are different representatives of the same coset. The definition of $F(a+\mathfrak p)$ depends on the representative $a$. To show well-definedness, you need to show that for a different representative $b$ we have $F(a+\mathfrak p) = F(b+\mathfrak p)$. Now, $b+a-b$ and $a$ are the same representatives (i.e. $b+a-b= a$), so that $F(a+\mathfrak p) = F(b+a-b+\mathfrak p)$ is trivial. But $b+a-b$ and $b$ are different representatives for the same coset, so that $F(b+a-b+\mathfrak p) = F(b+\mathfrak p)$ is to be shown. At some point, you have to make use of the actual definition of $F$.
Notice that $f(\mathfrak p) \subseteq \mathfrak q$ and hence, if $a+\mathfrak p = b+\mathfrak p$, then $b-a\in \mathfrak p$, i.e. $f(b-a)\in \mathfrak q$ and thus
\begin{align*}
F(a+\mathfrak p) &= f(a) + \mathfrak q = f(a) + f(b-a) + \mathfrak q\\
&= f(a+b-a) + \mathfrak q = f(b) + \mathfrak q\\
&= F(b+\mathfrak p).
\end{align*}
You could also apply the homomorphism theorem for rings to the composition $A\stackrel{f}{\to} B\to B/\mathfrak q$ (whose kernel is $\mathfrak p = f^{-1}\mathfrak q$), where $B\to B/\mathfrak q$ is the canonical projection.
A: You can try proof that a singular element of $A/p$ it has an only imagen on $B/q$. Let $a_1+p\,a_2+p \in A/p$ and suppose that $a_1+p=a_2+p$ then, $a_1-a_2\in p$ thus, $f(a_1-a_2)=f(a_1)-f(a_2)$ since $f$ is a ring homomorphism. Now, since $p=f^{-1}(q)$ there exist $k\in q$ such that $f(a_1)-f(a_2)=k$ and this implies that $f(a_1)+q=f(a_2)+q$ in other words, $F(a_1+p)=F(a_2+p)$, adn $F$ is well defined. 
A: I will attempt to turn my discussion with @DonAntonio in the comments into an actual answer. Since this is heavily influenced by discussion with another user, I have made this answer community wiki. In the course of this answer, I will be talking a lot about relations, rather than functions. At the end, I will say a few words about why we can legitimately skip out all this talk of relations.
When we say we say we define $F\colon A/\mathfrak{p}\to B/\mathfrak{q}$ by $F(a+\mathfrak{p})=f(a)+\mathfrak{q},$ we are not really defining a function here. We have not defined a function until we have shown that the "function" in question is well-defined. 
A possible save is to say we are defining a relation between the sets $A/\mathfrak{p}$ and $B/\mathfrak{q},$ a relation $R$ defined by $(a+\mathfrak{p})R(f(a)+\mathfrak{q})$ for all $a\in A.$ Proving "the function is well-defined" is the same thing as proving that "this relation is actually a function," just described in different words. To prove this relation is a function, we need to show that the set of elements in $B/\mathfrak{q}$ related to $a+\mathfrak{p}$ is a singleton set.
First, what does your argument show? If we interpret $F(a+\mathfrak{p})$ to mean "the set of elements related to $a+\mathfrak{p}$", then you have shown that the function mapping each element to its set of related elements is well-defined. This is clearly not the same as showing that the set of elements related to an element is a singleton set, so your argument does not prove that the function is well-defined.
How do we actually prove that $F$ is well-defined? That is, how do we prove that the relation is in fact a function? To do this, we have to use the definition of the relation in question. Suppose $a+\mathfrak{p}=b+\mathfrak{p};$ we will show that $f(a)+\mathfrak{q}=f(b)+\mathfrak{q}.$ The point is that if $x,y\in B/\mathfrak{q}$ with $(a+\mathfrak{p})Rx$ and $(b+\mathfrak{p})Ry,$ then for $R$ to define a function we should have $x=y$ if $a+\mathfrak{p}=b+\mathfrak{p};$ but the definition of the relation is that $(a+\mathfrak{p})R(f(a)+\mathfrak{q})$ and $(b+\mathfrak{p})R(f(b)+\mathfrak{q}),$ so it suffices to show that $f(a)+\mathfrak{q}=f(b)+\mathfrak{q}$ whenever $a+\mathfrak{p}=b+\mathfrak{p}.$
Now the actual proof is as follows:

Suppose $a+\mathfrak{p}=b+\mathfrak{p}.$ Then $a-b\in\mathfrak{p}$ (standard fact), so $f(a)-f(b)=f(a-b)\in\mathfrak{q}$ (by assumptions). Hence $f(a)+\mathfrak{q}=f(b)+\mathfrak{q}$ (the same standard fact).

By our previous remarks, this is all we need to prove that $F$ is well-defined, so now we can continue using $F$ as a function all we like. Note that the little argument in the pink box said nothing about general relations; that's why in general we don't need to say anything about relations. That technical stuff (which appears to me to be somewhat brushed under the rug) is all wrapped up in the preceding paragraph.

It is perhaps worth noting that the only real objection I have made is that you have used some notation before checking that the map is indeed a map; but, in my opinion, if this is such a trivial qualm that you can ignore it, then there is no point in explicitly proving well-defined-ness. Put another way, because well-defined-ness is something worth proving, this objection is a valid objection. After all, this is the reason we prove well-defined-ness in the first place.
