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(I hope this time it's less wrong)
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Let $T$ be a tensor field on the manifold $M$, $\nabla$ a connection, $v$ a tangent vector at $x\in M$, and $V$ a vector field such that $V(x)=v$.

Then the intuition is as follows:

The covariant derivative $\nabla_v T$ is the derivative of $T$ along a geodesic arc $\gamma$ for $\nabla$ which has direction $v$ at $x=\gamma(0)$. The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via parallel transport.

(Remark: here "geodesic arc" should be made more precise, as geodesics emanating from $x$ are determined as parametrized curves and it may happen that the geodesic in the direction $v$ doesn't have velocity $v$)


The (value at the point $x$ of the) Lie derivative $\mathcal{L}_VT$ is the derivative of $T$ along the flowline of $V$ (passing through $x$). The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via pullback along the local flow of $V$.


Edit. I happen to have re-read this old answer of mine, and I find that it was indeed misleading as indicated by Dean Yang in the comments. Let's see if it can be phrased better:

In both cases, we want to understand the derivatives as the velocity of curves in the same finite dimensional vector space $T_x M$ (or its tensor powers $T^{p,q}_x M$). How to do this?

  • In the case of $\mathcal{L}_V T$, we use the flow $\varphi_V^t$ of $V$.
    So the curve is $t\mapsto \varphi_{V,\star}^{-t}(T(\varphi_V^t (x)))$ (for contravariant tensors; for covariant ones or mixed ones, we use pullback along $\varphi_V^t$ instead, where needed).
  • In the case of $\nabla_v T$, we use the parallel transport $\Pi_{\eta,t}$ along $\eta$ with respect to $\nabla$. So the curve is $t \mapsto \Pi_{\eta,t}^{-1}(T(\eta(t)))$. (Here $\eta$ denotes any smooth curve passing through $x$ at $t=0$ with velocity $v$, and $\Pi_{\eta,t}:T_x M\to T_{\eta(t)}M$)
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