This is the exercise 12.2 of An Introduction to Manifolds of Loring Tu:
Let $(U,\phi )=(U,x^1,\ldots ,x^n)$ and $(V,\psi)=(V,y^1,\ldots ,y^n)$ overlapping charts on a manifold $M$. Then they induce coordinate charts $(TU,\tilde \phi )$ and $(TV,\tilde \psi )$ on the total space $TM$ of the tangent bundle with transition function: $$(x^1,\ldots,x^n,a^1,\ldots ,a^n)\mapsto (y^1,\ldots ,y^n,b^1,\ldots ,b^n)$$
(a) Compute the Jacobian matrix of the transition function $\tilde \psi \circ \tilde \phi ^{-1}$ at $\phi (p)$.
(b) Show that the determinant of the previous Jacobian matrix is $(\det[\partial y^i/\partial x^j])^2$.
To understand the notation: there $(a^1,\ldots ,a^n)$ is a representation of a tangent vector $v=\sum_{j=1}^n a^j\frac{\partial}{\partial x^j}\in T_pU$ and $(b^1,\ldots ,b^n)$ is a representation of the same tangent vector using the coordinate chart in $T_pV$ induced by $\psi $, that is, the coordinates $(a^1,\ldots ,a^n)$ and $(b^1,\ldots ,b^n)$ can be seen as functions on $T_p(U\cap V)$ that depends on the chosen $v$, and its easy to check that $b^k=\sum_{j=1}^na^j\frac{\partial y^k}{\partial x^j}$ and $a^k=\sum_{j=1}^nb^j\frac{\partial x^k}{\partial y^j}$.
Now, the Jacobian of $\tilde \psi \circ \tilde \phi ^{-1}$ have the form $$ \begin{align*} [\partial (\tilde \psi \circ \tilde \phi ^{-1})]&=\begin{bmatrix} \frac{\partial y^1}{\partial x^1}&\cdots &\frac{\partial y^1}{\partial x^n}&\frac{\partial y^1}{\partial a^1}&\cdots &\frac{\partial y^1}{\partial a^n}\\ \vdots &&\vdots &\vdots &&\vdots \\ \frac{\partial y^n}{\partial x^1}&\cdots &\frac{\partial y^n}{\partial x^n}&\frac{\partial y^n}{\partial a^1}&\cdots &\frac{\partial y^1}{\partial a^n}\\ \frac{\partial b^1}{\partial x^1}&\cdots &\frac{\partial b^1}{\partial x^n}&\frac{\partial b^1}{\partial a^1}&\cdots &\frac{\partial b^1}{\partial a^n}\\ \vdots&&\vdots &\vdots &&\vdots \\ \frac{\partial b^n}{\partial x^1}&\cdots &\frac{\partial b^n}{\partial x^n}&\frac{\partial b^n}{\partial a^1}&\cdots &\frac{\partial b^n}{\partial a^n} \end{bmatrix} \end{align*} $$ And $$ \frac{\partial b^k}{\partial x^j}=\frac{\partial}{\partial x^j}\sum_{\ell =1}^na^\ell \frac{\partial y^k}{\partial x^\ell }=\sum_{\ell =1}^na^\ell \frac{\partial^2 y^k}{\partial x^j\partial x^\ell }\\ \frac{\partial b^k}{\partial a^j}= \frac{\partial}{\partial a^j}\sum_{\ell =1}^na^\ell \frac{\partial y^k}{\partial x^\ell }=\frac{\partial y^k}{\partial x^j}+\sum_{\ell =1}^na^\ell \frac{\partial^2 y^k}{\partial a^j\partial x^\ell } $$
because $\frac{\partial a^k}{\partial x^j}=0$ for all $j$ and $k$, but I get stuck here, that is, I dont see how to evaluate these results at $\phi (p)$. Indeed $\phi (p)$ is not a point of the domain of $\tilde \psi\circ \tilde \phi ^{-1}$ so the exercise is not clear. Can someone help me?
$\tilde \psi \circ \tilde \phi ^{-1}(p,b)=(y^1(p),\cdots,y^n(p),\sum_{i=1}^nb_i\frac{\partial y^1}{\partial x^i},\cdots, b_i\sum_{i=1}^n\frac{\partial y^n}{\partial x^i})=(y^1(p),\cdots,y^n(p),J(\psi \circ \phi ^{-1})(p,b))b).$
Now, differentiate this. Since $J(\psi \circ \phi ^{-1})(p,b))$ is linear, it is its own derivative, so we get the block matrix we want:
$J(\tilde \psi \circ \tilde \phi ^{-1})(p,b))=\begin{pmatrix} J(\psi \circ \phi ^{-1})(p,b)) &0 \\ 0 & J(\psi \circ \phi ^{-1})(p,b)) \end{pmatrix}$