Is a linear map between two tensor spaces a tensor? For vector space $V$, suppose we have a linear map $U : V \rightarrow T^{1,1}V$, taking a vector and returning a tensor of type $(1,1)$ over $V$, i.e. a linear map $V \rightarrow V$.
We can feed $U$ two vectors and a dual vector and get back a scalar, just like a tensor of type $(1,2)$.  But is $U$ multilinear?
If that's true then, more generally, can we understand any linear map $T^{k,l}V \rightarrow T^{p,q}V$ as a tensor of type $T^{p+l,q+k}$?
 A: Yes. The easiest way to see it is to think of tensors as elements of 
$$ T^{(k,l)}V = (V^{*})^{\otimes k} \otimes V^{\otimes l}. $$
Now, a linear map $T \colon V \rightarrow W$ between finite dimensional vector spaces can be identified with an element of $V^{*} \otimes W$. In particular, a linear map $T \colon T^{(k,l)}V \rightarrow T^{(p,q)}V$ can be identified with an element of
$$ \left( T^{(k,l)} \right)^{*} \otimes T^{(p,q)} V = \left( (V^{*})^{\otimes k} \otimes V^{\otimes l} \right)^{*} \otimes \left( (V^{*})^{\otimes p} \otimes V^{\otimes q)} \right) \cong \\
 \left( \left( V^{*} \right)^{\otimes k} \right)^{*} \otimes \left( V^{\otimes l} \right)^{*} \otimes \left( V^{*} \right)^{\otimes p} \otimes V^{\otimes q} \cong \\
V^{\otimes k} \otimes \left( V^{*} \right)^{\otimes l} \otimes \left( V^{*} \right)^{p} \otimes V^{q} \cong \\
\left( V^{*} \right)^{\otimes (p + l)} \otimes V^{\otimes (q + k) } = T^{(p + l, q + k)} V.$$
where I used various isomorphisms such as 
$$(V \otimes W)^{*} \cong V^{*} \otimes W^{*}, \,\,\, \left( V^{*} \right)^{*} \cong V, \,\,\, V \otimes W \cong W \otimes V,$$
and so on.

This might look like I'm avoiding a direct answer to your question using "fancy isomorphisms" but the answer is actually there. For example, for your case, the resulting $(1,2)$ tensor, written in terms of $U \colon V \rightarrow \operatorname{Hom}(V,V)$ is given explicitly by
$$ (\varphi, v, w) \mapsto \varphi((U(v))(w)) $$
(or by $(\varphi, v, w) \mapsto \varphi((U(w))(v))$ which is not the same but gives you another way to convert $U$ into a $(1,2)$ tensor using a different isomorphism!). You can directly verify that this is multilinear in $\varphi,v,w$.
A: Take basis $e_\alpha$ in $V$. Then we can describe the map $U: T^{k,l}V\to T^{p,q}V$ as a table of coefficients:
$$
U(e_{\alpha_1}\ldots e_{\alpha_k}e^{\beta_1}\ldots e^{\beta_l}) =
\sum C_{\alpha_1\ldots \alpha_k\delta_1\ldots\delta_q}^{\beta_1\ldots \beta_l\gamma_1\ldots\gamma_p}e_{\gamma_1}\ldots e_{\gamma_p}e^{\delta_1}\ldots e^{\delta_q}
$$
Let's see what happens when we change basis from $e_\alpha$ to $f_\chi=\sum A_\chi^\alpha e_\alpha$:
$$
U(f_{\chi_1}\ldots f_{\chi_k}f^{\phi_1}\ldots f^{\phi_l})=
\sum D_{\chi_1\ldots \chi_k\omega_1\ldots\omega_q}^{\phi_1\ldots \phi_l\psi_1\ldots\psi_p}f_{\psi_1}\ldots f_{\psi_p}f^{\omega_1}\ldots f^{\omega_q},\\
\sum
A_{\chi_1}^{\alpha_1}\ldots A_{\chi_k}^{\alpha_k}
A_{\beta_1}^{\phi_1}\ldots A_{\beta_l}^{\phi_l}
U(e_{\alpha_1}\ldots e_{\alpha_k}e^{\beta_1}\ldots e^{\beta_l}) = \\
\sum
D_{\chi_1\ldots \chi_k\omega_1\ldots\omega_q}^{\phi_1\ldots \phi_l\psi_1\ldots\psi_p}
A_{\psi_1}^{\gamma_1}\ldots A_{\psi_p}^{\gamma_p}
A_{\omega_1}^{\delta_1}\ldots A_{\omega_q}^{\delta_q}
e_{\gamma_1}\ldots e_{\gamma_p}e^{\delta_1}\ldots e^{\delta_q}
$$
Then we use standard technique: for each factor $A_i^j$, we multiply left and right by $A^i_k$, contract index $i$ and then rename free index $k$ to $j$:
$$
A_i^j X_{\dots j}^{\ldots}=Y_{\ldots i}^\ldots,\\
A^i_k A_i^j X_{\dots j}^{\ldots}=A^i_k Y_{\ldots i}^\ldots,\\
\delta_k^j X_{\dots j}=A^i_k Y_{\ldots i}^\ldots,\\
X_{\dots j}=A^i_j Y_{\ldots i}^\ldots
$$
Thus,
$$
U(e_{\alpha_1}\ldots e_{\alpha_k}e^{\beta_1}\ldots e^{\beta_l}) = \\
\sum 
D_{\chi_1\ldots \chi_k\omega_1\ldots\omega_q}^{\phi_1\ldots \phi_l\psi_1\ldots\psi_p}
A^{\chi_1}_{\alpha_1}\ldots A^{\chi_k}_{\alpha_k}
A^{\beta_1}_{\phi_1}\ldots A^{\beta_l}_{\phi_l}
A_{\psi_1}^{\gamma_1}\ldots A_{\psi_p}^{\gamma_p}
A_{\omega_1}^{\delta_1}\ldots A_{\omega_q}^{\delta_q}
e_{\gamma_1}\ldots e_{\gamma_p}e^{\delta_1}\ldots e^{\delta_q}
$$
Comparing coefficients $C$ and $D$ we see that coefficients of map $U$ transform in tensor way during coordinate transformation. So maps $U$ are indeed tensor field $T^{(p+l,q+k)}$
