Notation of set that is the mapping of intersection I need to understand how to notate sets that comes from mapping with $f$. 
For $f:X\to Y$ and $A,B\subset X$ we have
$$
f(A\cap B)=\{f(x)\in Y:x\in A\land x\in B\}
$$
Is this true, or is the set notated otherwise?
 A: For $f~:~X\to Y$ and $A\subset X$ one has
$$f(A)=\{f(x)\in Y~:~x\in A\}$$
This is true regardless of how complicated of an expression $A$ happens to be.
For silly example, letting $A,B,C,D\subset X$, letting $(A\cup B)\cap (C\cup D)\cap (A\cup D) = Z$ one has
$$f(Z)=\{f(x)\in Y~:~x\in Z\}$$
or by replacing $Z$ with what it is equal to everywhere that it appears...
$$f((A\cup B)\cap (C\cup D)\cap (A\cup D)) = \{f(x)\in Y~:~x\in ((A\cup B)\cap (C\cup D)\cap (A\cup D))\}$$
In your specific case, noting that $A\cap B$ is itself a set you have of course $f(A\cap B)=\{f(x)\in Y~:~x\in A\cap B\}$.  You may opt to rewrite the condition $x\in A\cap B$ to be the condition $(x\in A)\wedge (x\in B)$ if you wish since these are equivalent to one another.
A: Yeah, that is right, although different books, courses or institutions MAY have slightly different conventions (for e.g., the use of square brackets). So it might be worth checking with your own institution. If you are writing a book then you define the convention! Some people would also describe that set as the image of $A \cap B$ under $f$.
A: What you wrote for $f(A\cap B)$ is true. Many texts and writers write $f(S)$ for $\{f(x): x\in S\}.$ This has a tacit assumption that no subset $S$ of the domain $X$ of $f$ is also a member of $X.$ And  I have yet to see this written when it was $not$ known nor assumed that $S\subset dom(f).$
It  is unnecessary (but not illogical) to include the  "$\in Y$" in what you have in your Q, because $f(x) \in Y$ iff $f(x)$ exists.
More simply you can also write $f(A\cap B)=\{f(x): x\in A\cap B\}.$
Set theorists write $\{f(x):x\in S\}=f''S$ (read $f$-double-prime $S$, or informally, $f$-double-tick $S$) when $S\subset dom (f).$ And they use brackets for dis-ambiguation. E.g. $f''(A\cap B)=\{f(x):x\in A\cap B\}$ because $f''A\cap B$ could be read as  $(f''A)\cap B.$
In set theory it is not rare for a member of a set to also be a subset of S. E.g. if $\emptyset \in S.$
