I have the following statement in my lecture:
Let $(M,g)$ be Riemannian manifold, $(U, \varphi)$ local chart on $U \subset M$, $\varphi=(x^1,...,x^n)$ and projection $\pi:TM \rightarrow M$. Then we have an induced chart
$(\pi^{-1}(U), \varphi_{*})$, where $\varphi_{*}=(x^1,...,x^n,y^1,...,y^n)$.
Now I don't know what the $y^i$ are supposed to be.
My interpretation is:
You can see $v \in TM$ as a pair $(p,v)$, if $v \in T_pM$. Then $\varphi_{*}((p,v))=(\varphi(p), v(x^1),...,v(x^n))$
Is that correct?
If you mean $v(x_i)$ as the tangent vector $v$ applied to the function $x_i$, then yes, you are correct.
Equivalently, this is the same thing as saying that a vector $v\in T_pM$ given in these coordinates by $(\varphi (p),y^1(p,v),\ldots,y^n(p,v))$ is the tangent vector
$$v=\sum_{i=1}^ny^i(p,v)\left(\frac{\partial}{\partial x^i}\right)_p$$
so you probably start using these coordinates without even thinking about it as soon as you prove that $T_pM$ is a vector space and that the $\left(\frac{\partial}{\partial x^i}\right)_p$'s form a basis.
To make it more clear as to why that statement actually means this, you have to know that $\varphi_*$ means the push-forward $TM\to T\mathbb{R}^n\cong\mathbb{R}^{2n}$. And the way this push-forward (or the push-forward of any other differentiable function between differentiable manifolds, for that matter) is defined is: if $v\in T_pM$ is tangent to the curve $c$ in $M$, then $\varphi_*v$ is tangent to the curve $\varphi\circ c$ (you can show that this is well-defined and that it's a linear transformation at each point $p$ between $T_pM$ and $T_{\varphi(p)}\mathbb{R}^n$). Well but $\left(\frac{\partial}{\partial x^i}\right)_p$ really is just notation for the vector tangent to the curve $c:(-\varepsilon,\varepsilon)\to M$ given in coordinates by $$(\varphi\circ c)(t)=\varphi(p)+(0,\ldots,0,t,0,\ldots,0)$$ where the $t$ is in the $i$-th coordinate, and this curve is obviously tangent to $e_i=(0,\ldots,0,1,0\ldots,0)$. So $\varphi_*\left(\frac{\partial}{\partial x^i}\right)_p=e_i$ and therefore
$$\varphi_*v=\varphi_*\left(\sum_{i=1}^ny^i(p,v)\left(\frac{\partial}{\partial x^i}\right)_p\right)=\sum_{i=1}^ny^i(p,v)\varphi_*\left(\frac{\partial}{\partial x^i}\right)_p=\sum_{i=1}^ny^i(p,v)e_i=(y_1(p,v),\ldots,y_n(p,v))$$
This shows that I was correct. To show that you were also correct, given all of this, is not hard:
$$v(x_i)=\left(\sum_{i=j}^ny^j(p,v)\left(\frac{\partial}{\partial x^j}\right)_p\right)(x_i)=\sum_{i=j}^ny^j(p,v)\left(\frac{\partial x^i}{\partial x^j}\right)_p=\sum_{i=j}^ny^j(p,v)\delta^i_j=y_i(p,v)$$