For the forward direction, there is a much more direct proof - mainly, a homotopy between a loop and a constant is a map from the disc already. It's in the spirit of your proof, but doesn't require the homotopy extension theorem.
So, let $f:S^1\rightarrow X$ and assume $f$ is null homotopoic. Let $F(x,t)$ be a homotopy between $f$ and a constant map with $F(x,1) = f$ and $F(x,0) = x_0$.
Here's the trick - think of $(x,t)$ coordinates as polar coordinates on the disc where $x$ corresponds to $\theta$ and $t$ corresponds to $r$. Alternatively, the space $S^1\times [0,1]/$~ where we identify $S^1\times\{0\}$ to a point is clearly homeomorphic to the disc, and it's easy to see that since $F(x,0) = x_0$ independent of the $x$ value, that $F$ descends to map on $S^1\times [0,1]/$~.
I will give an elementary proof of the problem using the fact, that $T^2$ is a topological group and that its universal cover is contractible. We will start with some constructions in homotopy theory of topological groups, which are required to understand the proof given below and added here for convenience.
First note, that $\mathbb R^2$ forms a topological group under addition: $(x,y) + (x',y') := (x+x',y+y')$ and that $\mathbb Z^2$ is a discrete normal subgroup thereof. We identify $T^2$ with the quotient of $\mathbb R^2$ by $\mathbb Z^2$, so that $T^2$ again becomes a topological group under addition. Moreover the quotient map $p: \mathbb R^2 \to T^2$ becomes a group homomorphism and is easily seen to be the universal covering projection ($\mathbb Z^2$ discrete subgroup $\Rightarrow$ $p$ is covering projection; $\mathbb R^2$ is contractible $\Rightarrow$ $p$ is universal).
Next, we observe, that for $[f],[g]\in [(X,\ast),(T^2,0)]$ the sum $[f]+[g] := [f+g]$ is well defined, turning $[(X,\ast),(T^2,0)]$ into a group. The same arguments show that $[(X,\ast),(\mathbb R^2,0)]$ is a group under point wise addition of representatives as well (as is $[(X,\ast),(G,1)]$ for any topological group $G$ with unit $1$), and that the map $p_\sharp : [(X,\ast),(\mathbb R^2,0)] \to [(X,\ast),(T^2,0)]$ given by $p_\sharp([f]) := [p \circ f]$ is a group homomorphism.
Now $\pi_1(T^2,0) = [(S^1,1), (T^2,0)]$ is a group in two ways, by means of composition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha \ast \beta]$ and by means of point wise addition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha + \beta]$.
Both operations share the same unit, the (class of the) constant loop sending everything to $0 \in T^2$ and denoted simply by $0: (S^1,1) \to (T^2,0)$. We can also observe, that for any loops $\alpha, \beta, \gamma, \delta$ we have $(\alpha + \beta) \ast (\gamma + \delta) = (\alpha + \gamma) \ast (\beta + \delta)$. Therefore $$[\alpha]+[\beta] = ([\alpha] \ast [0]) + ([0] \ast [\beta]) = ([\alpha] + [0]) \ast ([0] + [\beta]) = [\alpha] \ast [\beta],$$
hence the two operations are in fact the same on $\pi_1(T_2,0)$. The same argument can be used to show the analogous statement for $\pi_1(\mathbb R^2,0)$ (or $\pi_1(G,1)$ for any topological group $G$ with unit $1$).
Now back to the problem:
Given two maps $\varphi, \psi: T^2 \to T^2$, such that for some point $x \in T^2$, we have $\varphi(x) = \psi(x) = x$ and $\pi_1(\varphi) = \pi_1(\psi): \pi_1(T^2,x) \to \pi_1(T^2,x)$, we want to show $\varphi \simeq \psi$, where the homotopy can be taken relative to $x$. Replacing $\varphi$ with $\xi \mapsto \varphi(\xi + x) - x$ and $\psi$ with $\xi \mapsto \psi(\xi + x) - x$ if necessary, we may assume $x=0$. It will then suffice to show, that $\chi \simeq 0$, where $\chi := \varphi - \psi$.
Since the induced map $\pi_1(\chi): \pi_1(T^2,0) \to \pi_1(T^2,0)$ on fundamental groups is trivial (this is where we need all the constructions for topological groups), we can lift $\chi$ to a map $\bar{\chi}: (T^2,0) \to (\mathbb R^2,0)$ with $\chi = p \circ \bar{\chi}$.
We now define $H: T^2 \times I \to T^2$ by $H(x,t) = p(t\bar\chi(x))$, which is easily checked to be the required homotopy $0 \simeq \chi$.
Best Answer
No, the lecturer does not claim such thing. In fact he says something exactly opposite.
The most general definition of homotopy is the one you've provided. And in this sense two paths with different end points can be homotopic. In fact $[0,1]$ is contractible and therefore any two paths are homotopic (well, at least if the target space is path connected). Moreover the lecturer says the same thing. And the conclusion here is that this case is simply not interesting at all.
Then the lecturer defines "homotopy of paths" which is a term on its own, not just general homotopy connecting two paths. Yes, mathematics is a funny science where such nuances occure. Homotopy of paths is the usual homotopy with the additional condition that endpoints are fixed at every $t\in[0,1]$. And for that to work original paths have to share endpoints of course. In other words homotopy of paths is not even defined for paths with different endpoints, so the question "are those path homotopic" is meaningless.
And so there are two different definitions of homotopy in play here. And it turns out that this stronger notion is in many cases better, more useful.
Note that sometimes people use the term "free homotopy" for the general one, to emphasize that it does not fix points.
And remember to read theorems, proofs and claims carefully. Because the difference between homotopy and homotopy of paths seems small, but actually is very important.