Suppose we have finite-dimensional complex vector spaces $U$ and $V$, and we consider the projective spaces $X=P(U)$ and $Y=P(U\oplus V)$. The blowup $Y'$ is then
$$ Y' = \{ (L,M) \in P(U\oplus V) \times P(V) : L \leq U\oplus M\}. $$
The map $f:Y'\to Y$ is just $f(L,M)=L$. The exceptional divisor is
$$ X' = \{ (L,M) : L\in P(U), M\in P(V) \} = P(U)\times P(V). $$
Let $x$ be the first Chern class of $L$ in the evident sense, and let $y$ be the first Chern class of $M$. Put $n=\dim(U)$ and $m=\dim(V)$. Note that $Y'$ can be regarded as the projective bundle of the tautological bundle plus $U$ over $P(V)$, which determines its cohomology. We have

\begin{align*}
H^\ast(X) &= \mathbb{Z}[x]/x^n \\\\
H^\ast(X') &= \mathbb{Z}[x,y]/(x^n,y^m) \\\\
H^\ast(Y) &= \mathbb{Z}[x]/x^{n+m} \\\\
H^\ast(Y') &= \mathbb{Z}[x,y]/(x^n(x-y),y^m) \\\\
\end{align*}

The tangent bundle to $X=P(U)$ is $\text{Hom}(L,U)-1$, and using this we get $c(X)=(1+x)^n$ (or at least $(1\pm x)^{\pm n}$, depending on your conventions). Similarly $c(Y)=(1+x)^{n+m}$ and $c(X')=(1+x)^n(1+y)^m$. If we regard $Y'$ as a bundle over $P(V)$ then the horizontal tangent bundle is $\text{Hom}(M,V)-1$ and the vertical tangent bundle is $\text{Hom}(L,M+U)-1$ which I think gives $c(Y')=(1+x)^n(1+y)^m(1+x-y)$.

This gives most of what you want. I don't have time to write more just now.