None of the definitions implies the other:

The function $|x|^{2-m}$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ according to the second definition, but it is not according to the first one.

Conversely, the function $-1$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ in the sense of the first definition and it is not according to the second one.

If we assume that $u \geqslant 0$, then every $u$ superharmonic in $\mathbb R^m \setminus K$ for a compact $K$ is superharmonic in $\mathbb R^m \cup \{\infty\} \setminus K$ according to the second definition, so for *positive* superharmonic function, the first definition clearly implies the second one (but not vice versa).

I believe things become clearer if one writes both definitions in terms of the Kelvin transform $u^*(x) = |x|^{2 - m} u(|x|^{-2} x)$.

The second definition requires that $u^*$ is superharmonic in a neighbourhood of $0$, so in a sufficiently small ball $B_r$ we have
$$ u^*(x) = \int_{B_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$
for some $\mu \geqslant 0$, some $h$ harmonic in $B_r$, some $a \geqslant 0$ and some $b \in \mathbb R$.

On the other hand, the first definition requires that $u^*$ is superharmonic in a punctured neighbourhood of $0$, and $|x|^{m - 2} u^*(x)$ satisfies the "super-mean-value property at $0$", so in a sufficiently small ball $B_r$ we *should* have
$$ u^*(x) = \int_{B_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$
for some $\mu \geqslant 0$, some $h$ harmonic in $B_r$, some $a \in \mathbb R$ and some $b \leqslant 0$. Note: I might be wrong here, I did not have time to think about this carefully.

The difference is of course in the admissible range of $a$ and $b$.

*Edit: Some additional comments to the above item 3.*

(A) If $u^*$ is superharmonic in a ball $B_R$, then in a smaller ball $B_r$ we have
$$ u^*(x) = \int_{B_r} |x - y|^{2 - m} + h(x) $$
for some harmonic $h$; this is the Riesz decomposition theorem. If we take out the atom at $0$, then we can write
$$ u^*(x) = \int_{B_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$
for some $a \geqslant 0$, and this is equivalent to what I wrote above.

(B) Suppose now that $u^*$ is the Kelvin transform of a function superharmonic at infinity in the sense of the first definition. Then $u^*$ is superharmonic in $B_R \setminus \{0\}$, and since $u$ is bounded from below by a constant $-M$ in a neighbourhood of $\infty$, we have
$$ u^*(x) + M |x|^{2 - m} \ge 0 $$
in some ball $B_R$. This implies that $u^*(x) + M |x|^{2 - m}$ is superharmonic in $B_R$, not just $B_R \setminus \{0\}$. By the same argument as in (A) we find that
$$ u^*(x) = \int_{B_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$
for some harmonic $h$, but this time $a$ need not be nonnegative (we only know that $a \ge -M$, but $M$ can be arbitrarily large). This leads us to the expression given in the original answer, with an arbitrary real $b$.

Now why in fact we need $b \leqslant 0$? I did not check this carefully, but the average of $|x|^{m - 2} u^*(x)$ over a small ball $B_s$ seems to be equal to $a + c_m b s^{m - 2} + o(s^{m - 2})$ for an appropriate constant $c_m > 0$, and this is no greater than the limit $a$ only if $b \leqslant 0$.

The above shows that $b \leqslant 0$ is necessary. The next question is whether is is also sufficient. I guess it is, and this should be fairly easy to verify, but I have to stop here for the time being.