I am trying to understand covariant derivatives in GR.
So far, I understand that if $Z$ is a vector field, $\nabla Z$ is a $(1,1)$ tensor field, i.e. the map $\eta,X \mapsto \eta(\nabla_XZ)$ is linear in $\eta$ and $X$. The map $Z\mapsto\nabla_XZ$ is not a tensor field since, by the Leibniz rule, it is not linear in $Z$: $$\nabla_X(fZ)=(\nabla_Xf)Z+f(\nabla_XZ)$$ where $\nabla_Xf:=X(f)$. My lecture notes then state that the action of $\nabla$ is extended to general $(r,s)$ tensor fields by the Leibniz rule, giving an $(r,s+1)$ tensor field as a result. The example given is $\nabla \eta$, with $\eta$ a $(0,1)$ tensor field:$$(\nabla\eta)(X,Y)\equiv\nabla_X\eta(Y):=\nabla_X(\eta(Y))-\eta(\nabla_XY)$$ which is easily seen to be linear in $X$, while the first terms from the Leibniz rule cancel to give linearity in $Y$ as well, so this is a $(0,2)$ tensor.
They then suggest to try and define $\nabla T$ for $T$ a $(1,1)$ tensor. My attempt at this is: $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))+T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$ Again, linearity in $X$ is easy and the Leibniz rule gives linearity in $Y$. However, this doesn't seem to be linear in $\eta$. First, I find that: $$\begin{eqnarray} \nabla_X(f\eta+\omega) (Y)&=& \nabla_X((f\eta+\omega)(Y)) - (f\eta+\omega)(\nabla_XY)\\ &=& (\nabla_Xf)\eta(Y)+f\nabla_X(\eta(Y))+\nabla_X(\omega(Y))-f\eta(\nabla_XY)-\omega(\nabla_XY)\\ &=& (\nabla_Xf)\eta(Y) + f \nabla_X\eta(Y)+\nabla_X\omega(Y) \end{eqnarray}$$ hence $\nabla_X(f\eta+\omega) = (\nabla_Xf)\eta + f \nabla_X\eta+\nabla_X\omega$. Therefore: $$\begin{eqnarray} \nabla_XT(f\eta+\omega,Y)&=& \nabla_X(T(f\eta+\omega,Y))+T(\nabla_X(f\eta+\omega),Y)-T(f\eta+\omega,\nabla_XY)\\ &=& (\nabla_Xf)(T(\eta,Y))+f\nabla_X(T(\eta,Y))+\nabla_X(T(\omega,Y))+(\nabla_Xf)T(\eta,Y)\\&&+fT(\nabla_X\eta,Y)+T(\nabla_X\omega,Y)-fT(\eta,\nabla_XY)-T(\omega,\nabla_XY)\\ &=& 2(\nabla_Xf)(T(\eta,Y)) + f\nabla_XT(\eta,Y)+\nabla_XT(\omega,Y) \end{eqnarray}$$ where the term $2(\nabla_Xf)(T(\eta,Y))$ shows that $\nabla_XT(\eta,Y)$ is not linear in $\eta$. However, if I defined it as:$$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)+T(\eta,\nabla_XY)$$ instead, then the same calculation shows that this is linear in $\eta$, but no longer linear in $Y$. So I am not sure which, if either, of them is the correct definition.
UPDATE: Thanks to Michael Albanese's answer I now understand that the correct definition is $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$ which is indeed linear in $\eta$, $X$ and $Y$. For the benefit of others, I'll mention that my confusion arose because I was trying to reverse engineer the $+$ and $-$ signs from the coordinate basis formula:
$$\begin{eqnarray} T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s};\rho} =T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s},\rho}&+\Gamma^{\mu_1}_{\sigma\rho}T^{{\sigma}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}+\ldots+\Gamma^{\mu_r}_{\sigma\rho}T^{{\mu_1}\ldots{\sigma}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}\\ &-\Gamma^{\sigma}_{\nu_1\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\sigma}\ldots{\nu_s}}-\ldots-\Gamma^{\sigma}_{\nu_s\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\sigma}} \end{eqnarray}$$ But I now see how they follow from the correct definition. In the case of a $(1,1)$ tensor we have $T^{\mu}_{\phantom{\mu}\nu\,;\rho}:=\nabla_{e_\rho}T(f^\mu,e_\nu)$, where I am using $f^\mu$ and $e_\nu$ for $\mathrm{d}x^\mu$ and $\frac{\partial}{\partial x^\nu}$. Hence we have: $$\begin{eqnarray} \nabla_{e_\rho}T(f^\mu,e_\nu) :=& \nabla_{e_\rho}(T(f^\mu,e_\nu))-T(\nabla_{e_\rho}f^\mu,e_\nu)-T(f^\mu,\nabla_{e_\rho}e_\nu) \end{eqnarray}$$ The first and third terms are easily evaluated: $$\nabla_{e_\rho}(T(f^\mu,e_\nu))=e_\rho(T(f^\mu,e_\nu))=e_\rho(T^{\mu}_{\phantom{\mu}\nu})=T^{\mu}_{\phantom{\mu}\nu,\rho}$$ $$T(f^\mu,\nabla_{e_\rho}e_\nu)=T(f^\mu,\Gamma^{\sigma}_{\nu\rho}e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T(f^\mu,e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$ For the second term, since $f^\mu$ is a $(0,1)$ tensor field, $\nabla f^\mu$ is a $(0,2)$ tensor field and hence $\nabla_{e_\rho}f^\mu$, for fixed $e_\rho$, is a $(0,1)$ tensor field. We evaluate: $$\begin{eqnarray} (\nabla_{e_\rho} f^\mu)_\lambda=\nabla_{e_\rho}f^\mu(e_\lambda) &:= &\nabla_{e_\rho}(f^\mu(e_\lambda))-f^\mu(\nabla_{e_\rho}e_\lambda) \\ &= & e_\rho(\delta^\mu_\lambda)-f^\mu(\Gamma_{\lambda\rho}^\sigma e_\sigma)\\ &= & 0 -\Gamma_{\lambda\rho}^\sigma f^\mu( e_\sigma)\\ &= & - \Gamma_{\lambda\rho}^\sigma \delta^\mu_\sigma\\ &= & - \Gamma_{\lambda\rho}^\mu \\ &= & - \Gamma_{\sigma\rho}^\mu \delta^\sigma_\lambda\\ &= & -\Gamma_{\sigma\rho}^\mu f^\sigma( e_\lambda)\\ \end{eqnarray}$$ Therefore $\nabla_{e_\rho}f^\mu = -\Gamma_{\sigma\rho}^\mu f^\sigma$ and hence: $$T(\nabla_{e_\rho}f^\mu,e_\nu)=T(-\Gamma_{\sigma\rho}^\mu f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T(f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}$$ Thus, we have: $$\nabla_\rho T^{\mu}_{\phantom{\mu}\nu}=T^{\mu}_{\phantom{\mu}\nu,\rho}+\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}-\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$ as expected.
