Let $f:\mathbb{R}^n\to\mathbb{R}^m$ with $n > m$ be smooth and for each $x\in\mathbb{R}^n$ let $H_x(\cdot, \cdot)$ denote the Hessian third-order tensor of $f$. Basically given any two vectors $v, w\in\mathbb{R}^n$ it returns a third vector $$ H_x(v, w)\in\mathbb{R}^m. $$
Is this third-order tensor linear in both arguments? That is $$ H_x(a v_1 + b v_2, cw_1 + d w_2) = acH_x(v_1, w_1) + adH_x(v_1, w_2) + bcH_x(v_2, w_1) + bdH_x(v_2, w_2) $$ for any four constants $a,b,c,d\in\mathbb{R}$ and four vectors $v_1, v_2, w_1, w_2 \in\mathbb{R}^n$?
Bonus
Also, given n $n\times n$ matrix $A$ is it also linear in the sense that $H_x(Av, w) = AH_x(v, w)$?
There is a bit of competing terminology here. The Hessian operator $H$ is a linear map between functions, as per my comment. But the Hessian of $f$ is also a linear map on vectors (a tensor!). I understand now that that is what you are asking about.
A $(p,q)$ tensor is defined as a linear map $T:{\mathcal V^*}^p\times\mathcal V^q\to\mathbb R$. So linearity actually follows immediately from that. But I will illustrate.
Let's call the Hessian of $f$, $Hf$. As I said before, $$(Hf)^k{}_{ij}=\nabla_i\nabla_j f^k$$ For generic vectors $u,v$, $$\big((Hf)(u,v)\big)^k=(Hf)^{k}{}_{ij}u^iv^j$$ Which is clearly linear. Also
$$\big((Hf)(Au,v)\big)^k=(Hf)^k{}_{ij}(Au)^iv^j \\ =(Hf)^k{}_{ij}A^i{}_lu^lv^j$$ Which is, again, clearly linear.