After some thought, I believe that neither Fact 1 nor the Claim are actually true as stated. Fact 2 is true, but does not imply Fact 1. The argument against Fact 1 is simple: let $X = S^1 = \{z \in \mathbb{C} \mid \lvert z \rvert = 1\}$ be the circle, let $p \colon [0,1] \to S^1$ be $p(t) = e^{2\pi i t}$, and let $c \colon [0,1] \to \{x_0\} = (1,0)$. Then $p$ is very much homotopic to $c$, say via
$$H(s,t) = e^{2\pi i st};$$
we have $H(s,0) = c(s)$ and $H(s,1) = p(s)$. But $p$ is not homotopic to $c$ through loops, as Fact 1 claims. This is tricky to prove (how do you show that some homotopy doesn't exist?) and one would normally prove it with covering spaces, which show immediately that anything homotopic to $p$ has winding number $1$, while $c$ clearly has winding number $0$. I will simply leave it at that, since I don't think you can really do this with your current level of technology.
Note that any path is homotopic to any point along it, by a similar trick: just pick any $t_0 \in [0,1]$ and, since the interval is contractible, choose a continuous contraction $f(s,t)$ from the identity to $t_0$; for example,
$$f(s,t) = t_0 + s(t - t_0)$$
(at $s = 0$ you get $t_0$, at $s = 1$ you get $t$, and since $0 \leq s \leq 1$, we have $f(s,t) \in [0,1]$; in fact, $f(s,t)$ is between $t$ and $t_0$ for all $s$). Then given a path $p \colon [0,1] \to X$, the map $F(s,t) = p(f(s,t))$ is a homotopy from $p$ to $p(t_0)$. As the other answer says, homotopy of paths without fixing endpoints is pretty meaningless.
Anyway, the Claim is an even more specific fact, that $p$ and $c$ must be homotopic through loops based at $x_0$. This is impossible for the same reason.
I am actually a bit puzzled where you got Fact 1 from. I have a copy of the book Algebraic Topology by Greenberg and Harper, presumably a revision, and the proof in question is covered entirely, and without unproven claims, by just Fact 2 (their Lemma 3.3). I believe you may have misinterpreted the key step in the following expanded version of the argument there, which does produce a homotopy with the properties given in Fact 1, but from stronger hypotheses.
Theorem: If $X$ is contractible, then it is simply connected.
Proof: Let $p \colon [0,1] \to X$ be any loop with $p(0) = p(1) = x_0$. Because of this, we can define a new continuous function
$$q \colon S^1 \to X$$
where we consider $S^1 = [0,1]/(0 = 1)$. More explicitly, let $f \colon [0,1] \to S^1$ be the continuous map identifying $0$ with $1$, so we have $p = q \circ f$.
Suppose $X$ is contractible; then there is a homotopy
$$H(x,t) \colon X \times [0,1] \to X$$
with $H(x,0) = x$ and $H(x,1) = x_0$ from the identity function to the constant function $x_0$. From this, we define a map
$$H' \colon S^1 \times [0,1] \to X$$
by $H'(s,t) = H(q(s),t)$. This is a homotopy from $q$ to $x_0$, since
$$\begin{align}
H'(s,0) = H(q(s),0) = q(s) &&
H'(s,1) = H(q(s),1) = x_0
\end{align}$$
It is continuous, being a composition of continuous functions. Define a further homotopy
$$F(s,t) \colon [0,1]^2 \to X$$
by $F(s,t) = H'(f(s),t)$. It is continuous and a homotopy from $p$ to $x_0$, since
$$\begin{align}
F(s,0) = H'(f(s),0) = q(f(s)) = p(s) &&
F(s,1) = H'(f(s),1) = x_0.
\end{align}$$
$F$ has the property that
$$\begin{align}
F(0,t) &= H'(f(0),t) = H(q(f(0)),t) = H(p(0),t) = H(x_0,t) \\
&= H(p(1),t) = H(q(f(1)),t) = H'(f(1),t) = F(1,t).
\end{align}$$
That is, $F$ satisfies the criteria of Fact 1. (Note that we had to use the global homotopy $H$ to get $F$, not merely a homotopy from $p$ to $x_0$.) Defining $\alpha = \beta$ to be this common vertical-edge path, which is actually a loop at $x_0$ since $F(0,0) = F(0,1) = F(1,0) = F(1,1) = x_0$, we get from Fact 2 that $\alpha^{-1} p \alpha$ is homotopic to $x_0$ as loops based at $x_0$. That is, we have
$$[\alpha]^{-1} [p] [\alpha] = [x_0] = 1$$
as homotopy classes in $\pi_1(X, x_0)$. Moving $[\alpha]$ to the other side, we get $[p] = [\alpha][\alpha]^{-1} = 1$, so $[p]$ is the trivial element of the fundamental group. Since $p$ is arbitrary, $X$ is simply connected. $\square$
Based on the manipulations in this proof, the best approximation to Fact 1 that I can come up with is:
Proposition: Let $X$ be a topological space and $f, g \colon X \to X$ two homotopic continuous maps. Then for any loop $p \colon [0,1] \to X$ based at $x_0 = p(0) = p(1)$, there is a homotopy $F \colon [0,1]^2 \to X$ from $f\circ p$ to $g\circ p$, both loops at $x_0$, such that for each $t \in [0,1]$, the path $s \mapsto F(s,t)$ is a loop based at $x_0$; i.e. $F(0,t) = F(1,t)$ for all $t$.
Proof: In short, make $p$ into a map $q$ from the circle and compose the assumed homotopy from $f$ to $g$ with $q$ in the first variable. Then, unwrapping the circle, you find that the resulting homotopy from $f \circ p$ to $g \circ p$ has the property claimed. The details are above. $\square$
Best Answer
If $\varphi_t:S^1\times I\to X$ is the free homotopy between the loop $\varphi_0$ and the constant loop $\varphi_1\equiv x$, and $h$ is the path formed the images of $s_0$, i.e. $h(t)=\varphi_t(s_0)$, then define $h_t(s)=h(ts)$. At $t$ it traverses the path $h$ till the point $h(t)=\varphi_t(s_0)$, so it can be composed with $\varphi_t$, which itself can be followed by $\overline{h_t}$ to get back to $\varphi(s_0)$. So the product $h_t\cdot\varphi_t\cdot\overline{h_t}$ gives a bases homotopy between $\varphi_0$ and $h_1\cdot x\cdot\overline{h_1}$, the later being contractible.
The idea is also used in Lemma 1.19 of Hatcher's Algebraic Topology on page 37. Actually, this lemma is needed to show that homotopy equivalent spaces have isomorphic fundamental groups, so arguing that a point has trivial fundamental group would be a form of circular reasoning.