The basis vectors on a manifold are defined as partial derivative operators of any function that can be locally mapped around a point to $\mathbb R^n.$ The Christoffel symbols come about when a vector field on a such a manifold is differentiated, and the product rule calls for differentiating not just the components, but also these basis vectors, since they change from point to point.
Is there any sense, therefore, in which the Christoffel symbols can be understood as a second derivative?
No, they are not second derivatives of any kind. They're the "expansion coefficients" for differentiating one basis vector field along another. The expression $\nabla_XY$ means roughly to "differentiate the vector field $Y$ along the direction $X$". Now, one can consider any local frame $\{\xi_1,\dots, \xi_n\}$ of the tangent bundle, and in this regard, consider $\nabla_{\xi_i}(\xi_j)$. This is yet again another local vector field, so can be written as a linear combination $\nabla_{\xi_i}(\xi_j)=\Gamma^k_{ij}\xi_k$ for some $\Gamma$'s. So $\Gamma^{k}_{ij}$ tells us the $\xi_k$ component of the change in $\xi_j$ along $\xi_i$; but this 'change' arises due to a 'first derivative'. Indeed, the fact that $\nabla$ satisfies the product rule is very much characteristic of first-derivatives.
Do not be confused by the fact that the coordinate-induced basis vector fields can be written as $\frac{\partial}{\partial x^i}$. This merely has to do with one of the many equivalent ways one defines the tangent space to a manifold.
Edit In Response to Comments:
Ok my first sentence was too harsh (see Deane's excellent and concise answer for one interpretation, and below for a different special case interpretation), but still, I maintain that as far as it pertains to derivatives of vector field/sections of bundles (which is mainly why covariant derivatives are introduced), you shouldn't think of second derivatives.
Suppose $M$ is a smooth $m$-dimensional embedded submanifold of $\Bbb{R}^n$, let $\phi:A\to \phi[A]\subset M\subset\Bbb{R}^n$ be a local parametrization of $M$, where $A\subset\Bbb{R}^m$ is an open set (so $\phi$ is smooth, a homeomorphism onto its image, and has injective derivative at every point, all in the usual multivariable calculus sense). Now, for each $p\in \phi[A]\subset M$, and each $i\in\{1,\dots, m\}$ let us define \begin{align} \xi_i(p):=(\partial_i\phi)_{\phi^{-1}(p)}\equiv \frac{\partial \phi}{\partial u^i}\bigg|_{\phi^{-1}(p)}\in\Bbb{R}^n. \end{align} This is a vector tangent to $M$ at the point $p$; see this answer just in case you're not comfortable with relating the abstract definitions with concrete stuff in the special $\Bbb{R}^n$ case. Let $\langle\cdot,\cdot\rangle$ denote the standard inner product in $\Bbb{R}^n$, and define for each $a,b\in\{1,\dots, m\}$, \begin{align} g_{ab}(p):=\langle \xi_a(p),\xi_b(p)\rangle, \end{align} and let $[g^{ab}]$ denote the inverse matrix to $[g_{ab}]$. This $g$ is actually none other than the metric tensor on $M$ induced via pull-back of the standard RIemannian metric on $\Bbb{R}^n$. Now, by using the Levi-Civita connection on $M$, we can easily work out that the Christoffel symbols are \begin{align} \Gamma^k_{ij}(p)&=\frac{1}{2}g^{ks}(p)\left(\partial_i(g_{js}\circ \phi)_{\phi^{-1}(p)}+ \partial_j(g_{si}\circ \phi)_{\phi^{-1}(p)}- \partial_s(g_{ij}\circ \phi)_{\phi^{-1}(p)}\right)\\ &=g^{ks}(p)\cdot \langle(\partial_i\partial_j\phi)_{\phi^{-1}(p)}, (\partial_i\phi)_{\phi^{-1}(p)}\rangle\\ &\equiv g^{ks}(p)\cdot \left\langle\frac{\partial^2\phi}{\partial u^i\partial u^j}\bigg|_{\phi^{-1}(p)},\frac{\partial\phi}{\partial u^s}\bigg|_{\phi^{-1}(p)}\right\rangle \end{align} (the last equality is just a change to more classical Leibniz notation). So, ok the Christoffel symbols corresponding to the Levi-Civita connection for an embedded submanifold, and corresponding to a local parametrization, does depend on second derivatives of the parametrization.
Having said this, here's why I'd strongly urge you to not think that just because of this, the Christoffel symbols capture second derivatives
This relies on having $M$ be embedded in some $\Bbb{R}^n$, and we exploited heavily that this is an inner-product vector space (we needed the inner product of $\Bbb{R}^n$ and we needed the vector space structure to differentiate $\Bbb{R}^n$-valued maps, namely the parametrization $\phi$). Yes, there are several embedding theorems (even isometric ones), so in a sense, this is no loss of generality. However typically one needs much higher dimensional ambient space (and the larger the ambient space, the more seemingly superfluous information there is, which automatically makes it difficult to recognize intrinsic properties). We're also talking only about the Levi-Civita connection.
Elaborating on the previous comment, a general manifold is not given to us already embedded in $\Bbb{R}^n$, and as such it doesn't have "position vectors", or as I used above, $\Bbb{R}^n$-valued parametrizations $\phi$. These parametrizations aren't the fundamental quantities. i.e in basic math and physics courses, we learn about distances first, then we learn about position/displacement, and then we learn about speed and velocity and then about acceleration. However, in differential geometry, the fundamental object is the manifold, and its various tangent spaces (i.e velocities are primary object). "position vectors" are a meaningless concept unless you're already embedded. "speed" is an extra concept which can only be made sense once you have a metric tensor, and finally distances/lengths are obtained by integrating the speed of curves. So, our logical development is very much reversed (this is more general, and more useful). So, as much as possible, we should try to avoid these parametrizations when formulating the theory (for concrete calculations, we of course need the charts/parametrizations, but when developing the theory, they for many purposes obscure the picture).
Just because second derivatives of $\phi$ appear in this formula, it doesn't mean the $\Gamma$'s are "second order effects". Given any smooth function $f$, say $\Bbb{R}\to \Bbb{R}$, I can find a function $F_1$ such that $F_1'=f$. Likewise, I can find another function $F_2$ such that $F_2''=f$. FOr any positive integer, I can find a function which if differentiated that many times, yields $f$. So does that mean $f$ should be thought of as a "$0^{th}$ derivative" (since $f^{(0)}=f$)? Or should I think of $f$ as a "first derivative" (since $F_1'=f$)? Should I think of it as a "second derivative" (since $F_2''=f$)? Same thing with the $\Gamma$'s. (The $\nabla$ on the other hand satisfies the product rule, which is very much characteristic of first derivatives, and the $\Gamma$'s appear simply due to linear algebra, as the "expansion coefficients" relative to a basis).
The connections we're talking about here are more properly called connections on $TM$ (and the induced connections on the various tensor bundles $T^r_s(TM)$). A slightly more general situation is that you have a "vector bundle" $(E,\pi, M)$. VERY roughly speaking, this means we have two smooth manifolds $E,M$ and a surjective mapping $\pi:E\to M$. THe idea is that this is a "smoothly varying family of vector spaces". So for each $p\in M$, we have a vector space $E_p:=\pi^{-1}(\{p\})$. These vector spaces make up $E$, i.e $E=\bigcup_{p\in M}E_p$. Now, we can consider an "$E$-vector field", or more technically a smooth section of the vector bundle. This means we consider a map $\xi:M\to E$ such that $\pi\circ \xi=\text{id}_M$. Unwinding the definitions, this just means at each point $p\in M$ we have a vector $\xi(p)\in E_p$ in this particular vector space. Now, we can ask about what it means to take directional derivatives of such a mapping where the target space consists of not just a single vector space like $\Bbb{R}^n$, but many of them. Note that these vector spaces $E_p$ need not be the tangent spaces $T_pM$ (so the $\xi$'s need not have anything to do with $\frac{\partial}{\partial x^i}$). They could be something else entirely. There's also no $\Bbb{R}^n$-valued parametrizations lying around, so no second derivatives of parametrizations to be even considered. Yet, the question of such differentiation seems quite fundamental; we can define $\nabla$ in a straighforward way (Koszul's definition for a covariant derivative, or we can use the much more geometric definition of Ehresmann to eventually arrive at $\nabla$). And even in this general circumstance, the $\Gamma$'s can be viewed simply as a linear-algebraic consequence: they're just certain coefficients in the expansion of relative choices of bases.
The main point I wish to drive home is that sometimes, the way certain quantities look like when we generalize the theory tells us how we should interpret it. We have already encountered so many bumps in the road when trying to impart this second-derivative interpretation to the $\Gamma$'s. So, that should hopefully be strong enough motivation that it isn't a fruitful way to think about it.
The Christoffel symbols arise naturally when you want to differentiate a scalar function $f$ twice and want the resulting Hessian to be a $2$-tensor. When you work out the details, you discover that with respect to local coordinates, the Hessian of $f$ is given by $$ \nabla^2_{ij}f = \partial^2_{ij}f - \Gamma^k_{ij}\partial_kf. $$ In particular, if you set $f(x) = x^k$, you get $$ -\nabla^2_{ij}x^k = \Gamma^k_{ij}. $$
Also, notice that the Hessian of $f$ will always be a symmetric $2$-tensor if $\Gamma^k_{ij} = \Gamma^k_{ji}$, which is equivalent to saying that the connection is torsion-free. That's one of the reasons you want the Levi-Civita connection to be torsion-free.
Suppose we have a vector field depending on what point we are in space $v= v^i e_i$, the Christtofel appear naturally when we take the derivative of vector with the parameterization of the coordinate. They're sort of like a 'spatial derivative' of the basis vectors.
What do I mean by that? Suppose you move from one point in the surface to another, then the vector field on the surface can change due two reasons:
The change in basis part is what the Christoffel's try to capture.
Suppose I move from point to $A$ to point $B$, along a coordinate gridline:
$$e_i(B) - e_i(A) \approx \Gamma_{ik}^l e_l d \lambda$$
Note that I am tkaing that $B$ is the point at parameter value $\lambda+ d \lambda$ and $A$ is the point at parameter value $\lambda$ on the gridlines. For a particular value of $i$, we can think of RHS as a matrix multiplying on the unit vectors to say what the change in that basis is as we move along the gridline.
Also, I guess they are sort of second derivatives? In Pavel Grinfeld's Tensor analysis book, he talks about how given a parametrization of a surface we automatically know the tangent vector of the surface by differentiation of the position vector.
For example, let's consider a sphere:
$$ R(a,\theta,\phi)= a \sin \theta \cos \phi \hat{i} + a \sin \phi sin \theta \hat{j}+ a \cos \theta \hat{k}$$
If you differentiate the above vector w.r.t. the coordinates, we can get two tangents vector at a point i.e: $e_{\theta} =\frac{\partial R}{\partial \theta}$ and $e_{\phi} = \frac{\partial R}{\partial \phi}$. The Christoffel would then be related to the second derivative of position vector (going by previous eq which I introduced the symbols with).
$$ e_r= \frac{\partial R}{\partial r} = ( \sin \theta \cos \phi, \sin \phi \sin \theta, \cos \theta)$$
$$ e_{\phi} = a \sin \theta( - \sin \phi, \cos \phi)$$
$$ \frac{\partial e_r}{\partial \phi} = \sin \theta (-\sin \phi , \cos \phi)= \frac{e_{\phi} }{a}$$
This above equation is telling us about the Christoffel $\Gamma_{r \phi}^i$ for $ \{r , \theta ,\phi \}$ it tells us that $ r$ and $\phi$ entries are zero and $\theta $ entry is $\frac{1}{a}$
I guess the main relationship with the second derivative is the geodesic equation $\ddot{x}(t) + \Gamma(\dot{x}(t), \dot{x}(t))= 0$, or $\nabla_{\dot{x(t)}}\dot{x}(t) = 0$. I think the most intuitive explanation for the role of $\Gamma$ is for any curve $x(t)$, the acceleration $\ddot{x}$ is not tangent, and the term $\Gamma(\dot{x}(t), \dot{x}(t))$ is required to make it tangent. It is easy to see this in the example of the sphere. Consider a particle moving on the unit sphere, $x^Tx = 1$. Then $x(t)^T\dot{x}(t) = 0$, hence $$x(t)^T\ddot{x}(t) + \dot{x}(t)^T\dot{x}(t) = 0. $$ Thus, $x(t)^T\ddot{x}(t)\neq 0$ generally, or $\ddot{x}(t)$ is not tangent, as well-known from highschool physics. But the above equation implies $$x(t)^T\{\ddot{x}(t) + x(t)\dot{x}(t)^T\dot{x}(t)\} = 0 $$ or $\ddot{x}(t) + x(t)\dot{x}(t)^T\dot{x}(t)$ is always tangent. If we work with a vector field $b(t)$ along the curve $x(t)$, ie $x(t)^Tb(t) =0$ but not necessarily $b(t) = \dot{x}^T(t)$, we have $\dot{x}(t)^Tb(t) + x(t)^T\dot{b}(t) =0$, or $$x(t)^T\{\dot{b}(t) + x(t)\dot{x}(t)^Tb(t) \}=0 $$ hence $\dot{b}(t) + x(t)\dot{x}(t)^Tb(t)$ is always tangent. The term $$\Gamma(\dot{x}(t), b(t)) = x(t)\dot{x}(t)^T b(t)$$ gives us a connection, it is a Christoffel function. This $\Gamma$ is not the only adjustment to make $\dot{b}$ tangent, once we have $\Gamma$, we can add another bilinear term tangent to the sphere. Among all the possible adjustments $\Gamma$ to make $\dot{b} + \Gamma(x, b)$ tangent, there is a unique one compatible with the metric, the Levi-Civita connection. For the embedded sphere, the simplest adjustment $x(t)\dot{x}(t)^T b(t)$ is actually the Levi-Civita connection.
