Let $M$ be a smooth manifold of dimension $n$ and let $A\colon \underbrace{\mathfrak{X}(M)\times\ldots\times\mathfrak{X}(M)}_{\textit{r-copies}}\longrightarrow C^\infty(M)$ be a $(0,r)$-tensor field on $M$, with $r>2$.
Given a orthonormal basis $(e_k)_{k=1}^n$ of $T_pM,$ for all $p\in M$ we define the contraction at $p$, with respect to $i,j$ ($i<j$) as $$A(X^1,\ldots,X^{r-2})(p):=\sum_{k=1}^n A(X^1,\ldots,\underbrace{e_k}_{\textit{i-th}},\ldots, \underbrace{e_k}_{\textit{j-th}},\ldots,X^{r-2}).$$ It can be proved that this definition does not depend from the choice of the orthonormal basis $(e_k)$ of $T_pM$.
Q. How can we prove that as $p$ varies in $M$ the contraction at $p$ is a $C^\infty$-map?
Set $n= \dim M$.
In the toy model $r=2$, your tensor field is represented (with respect to any chosen local orthonormal frame $e_i$ of $TM$ around $p$) by a symmetric $n \times n$ matrix $a_{ij}(p)$ whose components are smooth functions of $p$. Contracting this tensor (in this case there is just one choice) just mean "fix $i=j$ and sum". Thus the contraction is the $(0,\,0)$-tensor (i.e, the function) $$a_{11}+a_{22}+ \cdots +a_{nn},$$ in other word it is nothing but the trace of the matrix. Since the entries are smooth functions of $p$, so is its trace.
In the case $r\geq3$ the argument is exactly the same. Let me write the case $r=3$, leaving the general case to you.
A $(0,3)$-tensor is locally represented by a trilinear form $$A(X, \, Y, \, Z)=\sum_{i, \, j, \, k=1}^n a_{ijk} \,x^i y^j z^k,$$ where the $a_{ijk}=a_{ijk}(p)$ are smooth functions of $p$, the $x^i$ are the coordinates of the vector field $X$ with respect to an orthonormal frame of $TM$ and similarly for the $y^j$ and the $z^k$.
We have now three possible contractions with respect to a pair of indices. For instance, contracting with respect to $(i, \, j)$ means "fix $i=j$ and sum on $k$", so we obtain the $(0,\,1)$-tensor $$A(Z) = \sum_k a_{11k}\,z^k+\sum_k a_{22k}\,z^k+ \cdots +\sum_k a_{nnk}\,z^k.$$ Thus the coefficient of $z^k$ is $a_{11k}+a_{22k}+ \cdots +a_{nnk}$, namely a smooth function of $p$ and the result follows.