Why values can not be replaced with their extensionally equal values in an intensional system?
Supposing we have $a$ and $b$ of type $A$ and $p : \mathrm{Id}_A(a,b)$, there simply is not any rule of type theory that would allow you to replace $b$ with $a$ arbitrarily. So one answer is "because type theory does not let you do that", and if you think that it can be done, please show me how. But I suppose that is a non-answer, what you really are asking is why is type theory designed in such a strange way.

There are many ways to understand equality. Students of mathematics are typically taught only one, namely the equality arising from set theory and classical mathematics. This kind of equality is very intuitive and easily understood. But as you learn more mathematics, you begin to understand that sometimes two objects are not really equal, but are "kind of equal". For example, they could be isomorphic, or connected by a path. So it makes sense to look for other explanations of equality that capture the idea of a more "relaxed" equality.

In mathematics it makes a lot of sense to say that functions are equal if they produce the same outputs, but this is not so in computer science. For example, if functions $f$ and $g$ produce the same outputs, but one is much faster than the other, then it is useless for a computer scientist to consider them equal.

Type theory is a formal system which allows us to express these ideas. The identity type $\mathrm{Id}_A(a,b)$ is not necessarily the "true equality" (the only one that a student of classical mathematics knows about, and therefore calls it "true") of $a$ and $b$, it could be one of these other notions of equality that you learn about when you study homotopy theory and computer science.

For instance, one way to undertand $p : \mathrm{Id}_A(a,b)$ is to say that $p$ is some sort of a "path" from $a$ to $b$. That is, $a$ and $b$ are not "really equal", but there is a way to get from $a$ to $b$. And so, if $P :A \to \mathrm{Type}$, we certainly would not expect $P a$ and $P b$ to be "exactly the same", but rather that "there is a way to get from $P a$ to $P b$". Indeed, this goes under name transport and it follows from the rules for the identity type: given $p : \mathrm{Id}_A(a,b)$, there is a isomorphism of types $\mathrm{transport}_p : P a \to P b$.