Suppose $M$ is a surface and suppose $X: U \subset \mathbb R^2\rightarrow M$ is a coordinate patch. Then for every $p \in X(U)$, the pair of vectors $(X_u(X^{-1}(p),X_v(X^{-1}(p) )$ is a basis of the tangent vector space $T_pM$. Let $$\pi_1:\bigcup _{p \in X(U)}T_pM\rightarrow \mathbb R, \quad \pi_2:\bigcup _{p \in X(U)}T_pM\rightarrow \mathbb R$$ be the function that associates to each tangent vector $v$ on $p$ the only real number $a$ such that $$aX_u(X^{-1}(p))+bX_u(X^{-1}(p)).$$ Hence, if $v_p$ is a tangent vector on $p$, it follows that $$v_p=\pi_1(v_p)X_u(X^{-1}(p))+\pi_2(v_p)X_u(X^{-1}(p)).$$
It's easy to see that $\pi_1$ and $\pi_2$ are $1$-forms for $M$, but I want to see that they are differentiable $1$-forms. i.e. I want to see that if $$V:X(u)\rightarrow\bigcup _{p \in X(U)}T_pM$$ is a differentiable vector field, then the function $\pi_1(V):X(U)\rightarrow \mathbb R$ is differentiable.
It suffices to show that $\pi_1(V)\circ X:U\rightarrow \mathbb R$ is differentiable.
That's where I get stuck.
If the vectors $X_u, X_v$ are always orthogonal, then it follows easy since $$\pi_1(V)\circ X=\frac{\langle X_u, V\circ X\rangle}{||Xu||}$$ and the internal product of two differentiable functions is differentiable, but I'm not managing to handle the general case. Can someone help me?
You can prove a vector field is smooth if and only if in local coordinates it is a smooth linear combination of the coordinate vector fields. This follows from the definition of the smooth structure on the tangent bundle and the definition of smooth mapping.
You are just taking the coefficients.