How would we show that for a tensor of any rank we can replace the partial derivatives by co-variant (Levi-Civita) derivatives, I was reading this is a GR text where it was left to the reader as an exercise.
As in
$L_{\xi}T^{a}_{b} = \xi^{c}\nabla_{c}T^{a}_{b} -T^{c}_{b}\nabla_{c}\xi^{a} + T^{a}_{c}\nabla_{b}\xi^{c}\ $
On
The question is not really clear, I hope I will (at least partially) answer it. Here is a coordinate-free approach which I found much more helpful than the local one.
Given a vector space $V$ a derivation on $V$ is a type-preserving endomorphism $D$ of the tensor algebra over $V$ satisfying Leibniz rule and commuting with contractions. Type preserving means that if $K$ is a tensor of type $(p,q)$, then $DK$ is still a tensor of type $(p,q)$. Leibniz rule amounts to say that $D(K \otimes L) = DK \otimes L + K \otimes DL$ for any pair $(K,L)$ of tensors on $V$. In particular, $D$ vanishes on tensors of type $(0,0)$.
For example, take a tensor $K$ of type $(0,2)$, that means $K$ eats two vectors. Denote contractions by $C$ (e.g. $C(X \otimes K) = K(X,\cdot)$) and observe they are linear. Then, since $D$ commutes with contractions and satisfies Leibniz rule \begin{align} 0 = D(K(X,Y)) & = D(C(Y \otimes C(X \otimes K))) \\ & = C(D(Y \otimes C(X \otimes K)) \\ & = C(DY \otimes K(X,\cdot) + Y \otimes D(C(X \otimes K))) \\ & = K(X,DY)+C(Y \otimes C(D(X \otimes K))) \\ & = K(X,DY) + C(Y \otimes C(DX \otimes K + X \otimes DK)) \\ & = K(X,DY)+K(DX,Y)+DK(X,Y). \end{align} This gives the formula $DK(X,Y) = -K(DX,Y)-K(X,DY)$.
Now, on a manifold the first term in this chain of equalities is not necessarily zero, because $K(X,Y)$ is a function, not a constant. This gives the more general formula \begin{equation} DK(X,Y) = D(K(X,Y))-K(DX,Y)-K(X,DY). \end{equation} You can play and find formulas for derivations of all kind of tensors - even if you do not know anything about tensor algebras, you can still try to repeat the steps above using the formal rules (Leibniz and contractions).
Both Lie derivatives $\mathcal{L}_X$ and covariant derivatives $\nabla_X$ are derivations, which means they behave exactly as $D$. In the particular case of a tensor field $T$ of type $(0,2)$ and Levi-Civita connection, you get \begin{align} \mathcal{L}_XT(Y,Z) & = \mathcal{L}_X(T(Y,Z))-T(\mathcal{L}_XY,Z)-T(Y,\mathcal{L}_XZ) \\ & = X(T(Y,Z))-T([X,Y],Z)-T(Y,[X,Z]) \\ & = X(T(Y,Z))-T(\nabla_XY-\nabla_YX,Z)-T(Y,\nabla_XZ-\nabla_ZX) \\ & = X(T(Y,Z))-T(\nabla_XY,Z)-T(Y,\nabla_XZ) \\ & \qquad + T(\nabla_YX,Z) + T(Y,\nabla_ZX) \\ & = \nabla_XT(Y,Z)+T(\nabla_YX,Z)+T(Y,\nabla_ZX). \end{align} If this does not mean much to you, write it in local coordinates: suppose the coefficients of the covariant derivative are given by functions $\Gamma$, namely $\nabla_i\partial_j = \Gamma_{ij}^k\partial_k$. Then \begin{align} (\mathcal{L}_kT)_{ij} & = (\nabla_kT)_{ij}+T(\nabla_i\partial_k,\partial_j)+T(\partial_i,\nabla_j\partial_k) \\ & = (\nabla_kT)_{ij}+T(\Gamma_{ik}^l\partial_l,\partial_j)+T(\partial_i,\Gamma_{jk}^l\partial_l) \\ & = (\nabla_kT)_{ij}+\Gamma_{ik}^lT_{lj}+\Gamma_{jk}^lT_{il}, \end{align} which should recall your expression (I guess that is for a tensor field of type $(1,1)$).
Hint Expand each term on the right-hand side in terms of Christoffel symbols. For example, $$\xi^c \nabla_c T^a{}_b = \xi^c (\partial_c T^a{}_b + \Gamma_{d c}^a T^d{}_b - \Gamma^d_{bc} T^a{}_d) .$$
After expanding all three terms, exactly the terms that contain Christoffel symbols cancel, leaving precisely the coordinate derivative formula for the Lie derivative $\mathcal{L}_\xi T$.
NB this computation uses that $\Gamma_{pq}^r$ is symmetric in $p, q$ but makes no reference to the metric, so in fact the statement is true for any torsion-free connection on a smooth manifold, not just the Levi-Civita connection on a Riemannian manifold.