Let $(M,g)$ be a Riemannian manifold of dimension $m$, and $(U,\varphi)$ local chart for $M$ with coordinates $x^1,\dots,x^m$.
Now, let $R=R_{ij} \ dx^i\otimes dx^j$ be a symmetric $2$-contravariant tensor on $M$, and define $R^2$ the tensor with components $(R^2)_{ij}=R_{ik}g^{kt}R_{tj}$, where $g^{-1}=g^{ij}\frac{\partial}{\partial x^i}\otimes\frac{\partial}{\partial x^j}$ inverse of $g$.
- Is $R^2$ globally defined on $M?$
I solved a similar exercise involving the trace $tr_g(A)=g^{ij}A_{ij}$ of a $2$-contravariant tensor $A=A_{ij} \ dx^i\otimes dx^j$ on $M$, by showing that $tr_g(A)$ is the trace of $A^{\sharp (i)}=A^{i'}_{j}\frac{\partial}{\partial x^{i'}}\otimes dx^j$, regarded as endomorphism of $T_p M$ through the isomorphism $T^{(1,1)}(T_p M)\cong End(T_p M)$. Thus, $tr_g$ is globally defined on $M$.
I tried to do something close with $R^2$, but I came up with nothing.
In the absence of a metric, it only makes sense to square $(1,1)$-tensors because they are pointwise endomorphisms of the tangent spaces. So if $A = (A_i^j)$ is a $(1,1)$-tensor, it makes sense to consider $A^2 = A \circ A$ whose components are $(A^2)_i^j = (A_i^k A_k^j)$.
If a metric $g$ is given, then you can use it to raise or lower indexes. You can now consider the $(0,2)$-tensor defined by $Q(X,Y) = g(A^2X,Y)$, whose components in coordinates are $$ Q_{ij} = g_{jk}(A^2)_i^k = g_{jk} A_i^{\ell} A_{\ell}^k. $$ Now, using the fact that $A_i^{\ell} = g^{m\ell}A_{im}$ and $g_{jk} A_{\ell}^k = A_{\ell j}$, one obtains that $$ Q_{ij} = g^{m\ell}A_{im} A_{\ell j} $$ which is basically your expression after relabelling the indexes. Since $Q$ is globally defined, so is the expression on the right hand side (whether $A$ is symmetric or not).