Let $M$ be a smooth manifold and $X_r^s:M\to T_r^s(M)$ is a section. Let $P\subseteq M$ be an open set and $X_1,\ldots,X_s\in\mathcal T(P)$ and $X^1,\ldots,X^r\in\mathcal T^*(P)$ where $\mathcal T(P)$ and $\mathcal T^*(P)$ are the sets of local smooth sections on $TP$(tangent bundle) and $T^*P$(cotangent bundle), respectively. I want to show that the function $X_r^s(X_1,\ldots,X_s,X^1,\ldots,X^r):P\to\mathbb R$ defined by $$X_r^s(X_1,\ldots,X_s,X^1,\ldots,X^r)(p)=X_r^s(p)(X_1^p,\ldots,X_s^p,X^1_p,\ldots,X^r_p)$$ is smooth if and only if section's component functions are smooth.
$$ \begin{align*} RHS &=C_{j_1\ldots j_s}^{i_1\ldots i_r}(p)dx_p^{j_1}\otimes\ldots\otimes dx_p^{j_s}\otimes\partial x_{i_1}^p\otimes\ldots\otimes\partial x_{i_r}^p(X_1^p,\ldots,X_s^p,X^1_p,\ldots,X^r_p)\\ &=C_{j_1\ldots j_s}^{i_1\ldots i_r}(p)(dx_p^{j_1}\otimes\ldots\otimes dx_p^{j_s})(X_1^p,\ldots,X_s^p)(\partial x_{i_1}^p\otimes\ldots\otimes\partial x_{i_r}^p)(X^1_p,\ldots,X^r_p)\\ &=C_{j_1\ldots j_s}^{i_1\ldots i_r}(p)dx_p^{j_1}(X_1^p)\cdots dx_p^{j_s}(X_s^p)\partial x_{i_1}^p(X_p^1)\cdots\partial x_{i_r}^p(X_p^r)\\ &=C_{j_1\ldots j_s}^{i_1\ldots i_r}(p)X_1^p(x^{j_1})\cdots X_s^p(x^{j_s})\color{blue}{\partial x_{i_1}^p(X_p^1)\cdots\partial x_{i_r}^p(X_p^r)} \end{align*} $$
Main Question: In the last line we have terms like $\partial x_{i_r}^p(X_p^r)$ and $\partial x_i^p(f)$ for $f\in C^\infty(p)$ is defined as $$D_i(f\circ x^{-1})(x(p))$$ Since $X^r\in\mathcal T^*(P)$, so $X^r$ is from $P\to T^*P$ and so $X_p^r$ is from $T_pP\to\mathbb R$ and so $X_p^r\notin C^\infty(p)$. Hence $\partial x_{i_r}^p(X_p^r)$ doesn't mean. What is wrong? Can anyone help me, please?
I will write the full proof for $X^1_1$ in your notation (just to not get cray with indices). So we have that locally: $$ X_1^1(p)=\sum_{i,j} C^i_j(p) dx^j_p\otimes \partial x_i^p$$
Assume the $C^i_j$ are smooth functions. Pick $X,\alpha$ smooth sections of $TM,T^*M$ respectively. Which locally we can write as $X=\sum_i x_i\partial x_i$ and $\alpha=\sum_i f_idx_i$, so $x_i,f_i$ are smooth functions as well. Then: $$ X_1^1(X,\alpha)(p)=\sum_{i,j} C^i_j(p)dx^j_p(\sum_hx_h(p)\partial x_h^p)\partial x_i^p(\sum_kf_k(p)dx^k_p)$$ now $dx_p^j(x_h(p)\partial x_h^p)=x_h\delta_{jh}$ and $\partial x_i^p(f_k(p)dx^k_p)=f_k(p) \partial x_i^p(x_k)=f_k(p)\delta_{ji}$ so the previous just becomes: $$\sum_{ij}C^i_j(p)x_j(p)f_i(p)$$ which is a sum and product of smooth functions and therefore smooth.
Now for the reverse direction we want to show that each $C^s_t$ is smooth if $X^1_1$ is. From the previous expression it suffices to pick $X=\partial x_t$ and $\alpha=dx_s$.