To see that the map $\tilde{\phi}$ is invertible note that
$$\tilde{\phi}^{-1}(x^1(p), \dots , x^n(p),v^1, \dots , v^n) = v^i \frac{\partial}{\partial x^i}|_p$$
this map is well-defined as the components of a vector are uniquely specified given a choice of coordinates (recall $v^i = V[x^i]$ for $V=v^i \frac{\partial}{\partial x^i}|_p$). Also, the coordinate chart $x: M \rightarrow \mathbb{R}^n$ is injective so there exists just one point $p$ to which the vector $(x^1(p), \dots , x^n(p))$ corresponds. So, I hope I have convinced you it is a bijection in the $x$ coordinates. Of course, if there exists a coordinate system $y$ also containing $p$ then $\tilde{\phi}$ written as
$$ \tilde{\phi}(w^i \frac{\partial}{\partial y^i}|_p) = (y^1(p),\dots, y^n(p), w^1, \dots , w^n) $$
is likewise a bijection. Indeed, we have two different local coordinate representatives. To show $\tilde{\phi}$ is smooth we should recall that as $x,y$ are compatible charts at $p$ the mappings $x \circ y^{-1}$ and $y \circ x^{-1}$ are smooth maps on $\mathbb{R}^n$ in the sense of multivariate real calculus on $\mathbb{R}$. Also, if the vector $V$ has components $v^i = V[x^i]$ in the $x$-coordinate system and $w^i=V[y^i]$ in the $y$-coordinate system then these are related due to the chain rule:
$$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$$
Observe,
$$ V=v^i\frac{\partial}{\partial x^i} = v^i\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j} = w^j \frac{\partial}{\partial y^j} $$
thus, equating coefficients of the $y$-coordinate derivations, the components of the vector $V$ at $p$ are connected by the so-called contravariant transformation rule:
$$ w^j = v^i\frac{\partial y^j}{\partial x^i}$$
Collecting our thoughts, we wish to argue $\tilde{\phi}$ is smooth by examining how two different coordinate representatives are related. Note:
$$ (x^1(p), \dots , x^n(p),v^1, \dots , v^n) \mapsto (y^1(p),\dots, y^n(p), w^1, \dots , w^n)$$
is a mapping on $\mathbb{R}^{2n}$ for which the Jacobian matrix has the form:
$$ \left[\begin{array}{cc} (y \circ x^{-1})' & 0 \\ 0 & [\frac{\partial y^j}{\partial x^i}] \end{array} \right]$$
Here we noted that $y = y \circ x^{-1} \circ x$ so differentiation of the mapping $x \rightarrow y$ at $p$ is in fact differentiation of $y \circ x^{-1}$ at $x(p)$. However, we know that derivative exists as the transition function $y \circ x^{-1}$ is smooth for charts containing a common point $p \in M$. Furthermore, just to be explicit, these two blocks in the Jacobian above are identical; $(y \circ x^{-1})' = [\frac{\partial y^j}{\partial x^i}]$. This is contravariance manifested, the components of a vector field change in the same fashion as the coordinates themselves. In any event, it should be evident by now that the Jacobian of the transition map for the $\tilde{\phi}$ map is nonsingular hence the transition map is smooth. Notice, if $M$ was a $C^k$ manifold then the transition maps $y \circ x^{-1}$ would just be $C^k$ and the tangent bundle would likewise be $C^k$. The map we considered in this discussion is sometimes called the adapted coordinate chart on the tangent bundle. A nice summary of it is that $\tilde{x} = (x,dx)$ in the sense that $\tilde{x}(p,v) = (x(p),dx_p(v))$. In that notation it is easier to argue $\tilde{y} = (y,dy)$ and so
\begin{align} \tilde{y} &= (y,dy) \\
&= (y \circ x^{-1} \circ x, d(y \circ x^{-1} \circ x)) \\
&= (y \circ x^{-1} \circ x, d(y \circ x^{-1}) \circ dx)) \\
&= (y \circ x^{-1}, d(y \circ x^{-1})) \circ (x,dx) \\
&= (y \circ x^{-1}, d(y \circ x^{-1})) \circ \tilde{x} \\
\end{align}
once again demonstrating that $\tilde{x}$ and $\tilde{y}$ are compatible charts on $TM$ near $p$.
