(This is an elaboration on the comment of MG. I know I benefited a lot as an undergraduate from being shown this sort of argument once instead of having been told to check things in local coordinates, so I thought I'd do the same for you.)

The fact that $\nabla$ is a connection means that for every function $f$ $$\nabla_{fX}(Y)=f\nabla_X(Y)$$ and $$\nabla_X(fY)=f\nabla_X(Y)+L_X(f)Y.$$ In other words, $\nabla$ it is linear over the ring of functions in the argument $X$, but not linear in $Y$, and instead satisfies some sort of a Leibniz rule for differentiating products. That non-linearity in $Y$ is precisely an obstruction for being a tensor. (There is nothing to check here: a tensor on a manifold is a linear over the ring of functions construction depending on several vectors and covectors, or, more eloquently put, a section of a tensor construction applied to the tangent and cotangent bundle! Saying that a tensor is something that transforms correctly in every coordinate system is such a mean thing to say to innocent students.)

What you instantly see is that the "correction term" $L_X(f)Y$ does not depend on $\nabla$, so when you compute the difference of two connections, it will disappear, and hence that difference will be linear in both $X$ and $Y$.

Finally, for two connections $\nabla^{(1)}$ and $\nabla^{(2)}$, the tensor $A(X,Y)=\nabla^{(1)}_X(Y)-\nabla^{(2)}_X(Y)$, being a linear operator in $Y$ for every fixed $X$, satisfies $A(X,0)=0$, hence the claim you want.