Let $f:A \rightarrow B$ be a ring homomorphism and $\mathfrak{a}_1,\mathfrak{a_2}$ be ideals of $A$. Let $\mathfrak{a}^e$ denote the extension of an ideal $\mathfrak{a}$ of $A$ in $B$. An exercise shows that $$(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e \subseteq \mathfrak{a}_1^e \cap \mathfrak{a}_2^e.$$
It seems that $(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e =\mathfrak{a}_1^e \cap \mathfrak{a}_2^e$ is wrong in general, but I can not find a counter-example.
On
Here is a general method for producing counter-examples in algebra that often works.
Let us try to find a universal counter-example to your statement where the ideals $\frak a_1$ and $\frak a_2$ are principal. The universal ring having $2$ principal ideals is the ring $$ A=\mathbf Z[x,y] $$ with the principal ideals ${\frak a}_1=(x)$ and ${\frak a}_2=(y)$. Since $A$ is a UFD, one has ${\frak a}_1\cap{\frak a}_2=(xy)$. One wants to construct a morphism of rings $f\colon A\rightarrow B$ such that $(xy)^e\subsetneq (x)^e\cap(y)^e$ in $B$. This strict inclusion means that there is a multiple of $x$ in $B$ which is also a multiple of $y$ in $B$, but which is not a multiple of $xy$ in $B$.
Let $B$ be the ring one obtains from $A$ by adjoining an element $z$ such that the multipe $zx$ is equal to the multiple $zy$. More precisely, let $$ B=\mathbf Z[x,y,z]/(zx-zy). $$ Let $f\colon A\rightarrow B$ be the natural morphism. Everything has been devised such that $$ zx=zy\in(x)^e\cap(y)^e. $$
Let us prove that $zx\not\in (xy)^e$. This can be done in several ways. Here is one. It suffices to show that $zx$ is nonzero in the quotient ring $$ C=B/(xy)=\mathbf Z[x,y,z]/(zx-zy,xy). $$ We can even take a smaller quotient and show that $zx$ is nonzero there: $$ D=C/(x-y)=\mathbf Z[x,y,z]/(x-y,xy)=\mathbf Z[x,z]/(x^2)=(\mathbf Z[x]/(x^2))[z] $$ The image of $zx$ in $D$ is obviously nonzero. It follows that $zx\not\in (xy)^e$.
Hint. Try a ring extension which is not flat (see here why), e.g. $K[X^2,X^3]\subset K[X]$. (I leave you the pleasure to choose the right ideals.)